Trigonometric Equations Multiple Angles 0 to 2pi Restriction

Trigonometric Equations Multiple Angles 0 to 2pi Restriction

BAM!!! I’m Mr. Tarrou. Today we are going
to take a look at solving trig equations, we are continuing that idea only this time
they are going to have multiple angles. We are going to be looking for all solutions
between zero and 2pi like I did in my previous lesson. Only again, we are looking at multiple
angles. What does that going to look like? Well, here is the first example. We have got
2 times cosine of 2x minus 1 equals 0. We want to solve this for all x values between
zero and 2pi. Well, if this material has not been taught to you yet and you are just trying
to go through your book and you are working these out, you are looking at this and going
ok I want all my answers between zero and 2pi. You hopefully think, ok I need to find
the angle measure and you get 2x by itself. That means we are going to add both sides
by one and divide both sides by two to undo that multiplication of two. Now we have the
cosine of 2x equal to one-half. Now please remember this is a trig function and it is
not just a variable. It is not cosine times two times x. We cannot just simply divide
away this two, it is the cosine OF 2x. I like to keep my angle measures wrapped up inside
a set of parenthesis. This is not mathematically necessarily, but it really highlights that
it is the math function of cosine being applied to this angle measure. Now, you want that
cosine function to get away from the angle measure of 2x, this double angle. You want
to undo the cosine function by doing the inverse, so we have 2x equals the inverse cosine of
1/2. Don’t forget we put the sides of a triangle, in this case cosine being x over r, when we
put the sides of those triangles into the trig function we get out an angle measure.
Again, if this is a brand new question and you have never done these before, you might
be thinking well ok again with this zero to 2pi, where is cosine equal to 1/2…or what
is the answer from the inverse cosine of 1/2. That is 2x is equal to ok…cosine is equal
to 1/2 at pi/3 and (so pi/3, 2p/3, 3pi/3, 4pi/3…) oh that is right 5pi/3. You might
be thinking, well ok that is the two places on one rotation of the unit circle where cosine
is equal to 1/2 and you finish the problem by multiplying both sides of this equation
by 1/2. We are not looking for trying to solve for the angle measure of 2x, we are trying
to solve for x. We get x is equal to one-half of pi/3 is pi/6 and 1/2 of 5pi/3 is 5pi/6.
So, the direction said solve for x and I want all the possible answers between zero and
2pi, and this is 30 degrees and this is 150 degrees. I have not even made it half way
around the circle yet. There might be more possible answers and indeed there is. See,
the deal is, if you want all of your possible answers what you have to focus on here is
this double angle. This 2x, or the multiple that is in front of the variable that you
are looking for, the x, that multiple that is within the angle that you are trying to
solve, you can think of this as how many times you need to go around the unit circle. I just
went around the unit circle like we normally have ever done when we deal with trig functions,
but when I took that last step of dividing both sides by two to get x alone we actually
got some relatively small angle measures. They are correct, but they are not all of
them are within zero to 2pi. Now, it says that x needs to be between zero and 2pi, not
2x. We are not looking to do the inverse cosine of 1/2 around the unit circle just once, all
the answers that around that first rotation of the unit circle, but we are going to have
to go around twice to get all of the possible answers. So now you are bring in the old concept
of coterminal angles. We have already got the two places on the unit circle where cosine
is equal to 1/2, but where are the other two? Well, one way to figure this out is to simple
count out again. Our first answer was at pi/3, so I can just make another full rotation and
land back again at pi/3, so we have got one, two, 3pi/3, four, five, six, and 7pi/3. One
of the other answers is going to be 7pi/3. Now if you have a really large multiple and
maybe just swing your arm around or put a bunch of tick marks on your paper, the only
other option to find another coterminal angle is adding by 2pi. If I want my fourth possible
answer and just show you another way of doing this, you have to come off of the side to
your scratch paper and do what is 5pi/3 plus another full rotation of 2pi. You are going
to need a common denominator, so I am going to multiply the top and bottom by three and
we get 5pi/3 plus 6pi/3. So my other possible answer there now that we have a common denominator
is 11pi/3. We have again the idea, the new idea, of solving these equations with trig
functions when you have a multiple angle, is just to remind yourself that the multiple
on the angle is simply telling you how many times you need to go around the unit circle
so that when you are done you have all of the possible answers that are required based
on the directions. Now we have four possible answers from going around the unit circle
twice, and 7pi/3 and 11pi/3 are bigger than 2pi but I have to finish my last step which
is multiplying both sides of this equation by 1/2. Now we have got the pi/6, we have
1/2 times 5pi/3 which is 5pi/6, and then 7pi/6, and 11pi/6. This isn’t always going to happen,
but all o these answers here are on the unit circle and we are comfortable that all of
those are definitely less than 2pi. Again one last time before we do our last three
examples, with multiple angles that tells you how many times to go around the unit circle
to find all of your answers. A couple more examples. The orange chalk is not wanting
to erase very good. We have tangent of 2x, excuse me 3x/2 is equal to negative square
root of three. This one is already set up. The only oddity is this 3x/2. Well 3.2 is
1.5, so we are going to go around the unit circle one and a half times to get all of
the possible answers that we want. 3x/2 is equal to the inverse tangent of negative square
root of three. Tangent is negative two and four. What first answer do we have, what first
angle that we have in our first one and a half rotations that puts in quadrant two.
Where is tangent equal to negative square root of three. If you remember your unit circle
and certainly by now you have, at 2pi/3 your coordinate is (-1/2, sqrt(3)/2) and when you
take that y over x the ratio of tangent the y value of square root of three over two over
negative one-half does cancel out to be negative square root of three. Our answers here are
going to be 3x/2 is equal to the first answer of 2pi/3, and then with that first one and
a half rotations we are getting again from the multiple of three-halves…three divided
by two…we have got 2pi/3, 3, 4, 5pi/3, 6, 7, 8pi/3. We don’t need to go around two full
rotations because again my multiple is three-halves or 1.5, so we need to go around the unit circle
one and a half times and these are my three possible answers. So, our last step is to
get that x value alone which means that we are going to need to multiply both sides of
this, it is not really an is just a list of answers (but it is because
of the=sign). Multiply both sides by two-thirds, that cancels out and we get x is equal to
2/3 times 2/3 is …yes 2/3… is equal to 4pi/9, and two times five
is 10pi/9, and two times eight is 16pi/9.
There we go. Our three answers between zero and 2pi for the tangent of 3x/2 equals negative
square root of three. Alright, let’s go on. For our next example, let’s take a look at
the cosine of 2x minus pi/4 equals square root of two over two. Alright, so this is
kind of a funny looking angle of 2x-pi/4. I think we are going to need to go around
that unit circle two times and let’s just see how the negative pi over four works out.
Well, we are going to have to get the cosine function away from the angle measure as always.
2x-pi/4 equals the inverse cosine of the square root of two over two. That means that we are
going to have 2x minus pi/4, and the inverse of cosine of square root of two over two…
Where on the unit circle is cosine equal to this ratio? …that would be pi/4, 7pi/4,
how about pi/4 plus 2pi to get that next rotation in, and 7pi/4 plus 2pi. That is going to give
us 2x-pi/4 equals pi/4, 7pi/4, finding a common denominator here showing you other ways to
work these out and again just re-emphasizing that when you go one full rotation away that
is the idea of coterminal angles and that is when trig functions are equal, we have
8 plus 1 is 9pi/4, and finally we have got 8 plus 7 is 15pi/4. Now let’s deal with this
pi/4 that is over here. I have got to move it over by adding pi/4 to all of these answers,
so I am adding by pi/4. I am also running out of space. We are going to have 2x is equal
to 1pi/4 plus 1pi/4 is 2pi/4, 7pi/4 plus 1pi/4 is 8pi/4, 9pi/4 plus 1pi/4 is 10pi/4, and
pi/4 plus 15pi/4 is 16pi/4. Ok, so our last step is to divide everything by two, or multiply
both sides of this list of answers…this equation…multiply everything by 1/2. Look
at what we have. We have x is equal to 2 divided by 2 so pi/4, 8 divided by 2 is four and 4/4
is 1pi, then we have 10 divided by 2 is 5pi/4, and 16 divided by 2 is 8 and 8 divided by
4 is 2pi. Now, remember that negative pi over four there? Because that minus pi/4 is there,
look to make sure that all of your final answers are actually within the restriction given
to you in the directions. Of course I have 2pi, it is in the interval but I have a rounded
parenthesis which means that I want to go up to 2pi but not actually include it. So,
that subtraction there is causing us to make sure to double check our answers and we went
just a little bit too far and I got one extra answer that I didn’t need or want. One last
example and we will be finished with this lesson. Let me double check and make sure
that I have that right. Yep. One last example. This example is going to get me a lots and
lots of answers. Let’s make sure that I have got enough room here. Here we go. Here we
go, here we go, here we go again…ooh probably should not sung that. The sine of 2x plus
sine of 4x is equal to zero. Ok, we have to solve this. How are we going to do that? Because
we also have multiple angles here but they don’t even match and that is a problem. This
is effectively like having two variables in the same equation. Unless we can factor those
apart or do something with the identities, this is not going to be solvable. If I use
the double angle identity for sine to reduce this down, I may have something that I can
actually solve. We have the sine of 2x, and the sine of 4x, I am going to use this identity…
The sine of 2x is equal to 2 sine(x) cosine(x) It is not so important that this is 2x and
this is 1x, what is important is that this angle is double what these two angles are,
so I have a four here instead of a two. That just means I am going to have 2x here. The
sine of 4x is equal to 2 times sine of 2x times cosine 2x equals zero. Now we still have these two variables but they will factor apart and that
means that we can work these out. These two terms both have a sine of 2x that is going
to factor out. The sine of 2x divided by the sine of 2x is one, plus the sine of 2x over
the sine of 2x cancels out leaving us with cosine of 2x equals zero. I am going to set
each one of these factors equal to zero. Solve each one. Where is sine equal to zero and
again we are talking about a double angle so we are going to go around the unit circle
twice. That is going to be at zero, pi, 2pi, and 3pi. Again, zero, pi, 2pi, 3pi, if I continue
to go all the way around I am going to get to 4pi but that is going to be too big because
of the restrictions that I erased. The answers need to be from zero and up to but not including
2pi. If you put an extra one in there, you are going to notice that it is too large when
you go back at the end and you check your answers. We are going to subtract both sides
by one, divide everything by two, again inverse trig function so 2x is equal to the inverse
cosine of negative one-half. Again with going around the unit circle twice that is going
to be 2x is equal to the inverse cosine of negative one-half, that is 2pi/3, 4pi/3, and 8pi/3, and finally 10pi/3.
WHOO! Running out of breath here. Multiply everything by a half. We get zero, pi/2, pi,
3pi/2. From this factor we get 1/2 of 2pi/3 is pi/3, and 1/4 of 4 is 2 so 2pi/4, and 4pi/3,
and 5pi/3. Just because I am a little bit paranoid, I am going to double check my answer
before I tell you guys ok. BAM!!! I am Mr. Tarrou. Go Do Your Homework:)


  • ProfRobBob says:

    I don't have a bunch of examples with bearing but the example starting at around minute 15 of Applications of Law of Sines and Cosines is a bearing question. I want to be more help, but I am teaching Calculus live for the first time as well as 3 other subjects so I am trying very hard just to stay ahead of my Calculus class. I hope my example is enough to help.

  • Brent Nachison says:

    do you have a video that explains how to solve problems like cos2x+sinx=1????

  • ProfRobBob says:

    I don't believe I have a video with that example. Is that cos(2x) or (cos(x))^2? You will want to start by writing the equation all in terms of one trig function and with the same angle measure. If a double angle, start with a double angle identity for cosine that only has the sine function on the right side of the identity.

  • Brent Nachison says:

    the first one and im confused

  • Ryan Macintyre says:

    You're such a James Spader look-a-like in this video, I love it haha 🙂 keep up the great work Mr. Tarrou

  • Lyyn says:

    you sir, are wonderful 

  • Jay Oberlander says:

    How do I tell the difference between using multiple angle and double angle ? Should x be theta for multiple angle ? 

  • Yeo Gyoung Kwak says:

    i love you

  • Foofoocuddlypoops19 says:

    Really wish you were my trig teacher. I'm doing so horrible right now and I'm almost done with the class. Here's hoping these videos for the next lessons will help and help me pass with a C  >.< 

  • Elizabeth Mulvey says:

    Thank you for taking the time to explain things in a way that actually makes sense! Keep the videos coming… Could you please also make some General Chemistry videos? lol

  • Denyese Gonsalves says:

    I believe you are a descendant of Archimedes. There is no other explanation. Wow thank you thank you thank you. I am on spring break and I have officially looked at more than 60 of your videos.

  • Adon Zulu says:

    You are the best and may God bless you.Thank you

  • IdentityRS says:

    Please help, I really have no Idea where you got the 5 pie over 3 from in the first place, and in general, how you get the main radians within the first pie range of all trigonometric equations. 

  • simplyximei says:

    A million times thank you!! You have no idea how much your videos have helped me. My math teacher doesn't give me the time and help that I need, but your videos have. Honestly, I don't think I'd be passing math without you. God bless!

  • Ross Chavelas says:

    Finally I find u on Facebook!!! Thanks for great videos…..

  • Nelson Villafane says:

    Thanks, you made this simple. The text book made it seem more complicated than it really is.

  • Oliver Josue says:

    you have serious sweg

  • J Sharp says:

    Thanks for using the phrase" Cos of 2x"………..just hearing you say "of " made a huge impact in trying to solve trig equations…..JJ

  • J Sharp says:

    You;re a good man Mr T…………….Thank you—–JJ

  • dickbonerman says:

    My pre-calc teacher has a class of 300+ people so getting him to elaborate on how to solve these problems during class is a nightmare, your videos are the reason I've been able to do well on my tests, these videos are lifesavers!

  • Eva T says:

    Hi, I just started to watch your videos and I find it to be very helpful. It has helped me with most of my Trig. questions that my instructor can't seem to explain to where I can understand how to get the solution or how the procedure works.  I do have a problem I would like to know how to get the solution for.  Cos theta= 1/4. I have gotten as far as finding the sides of the triangle and it (triangles) are in Quadrant I and IV. May I ask what I am missing? 

  • jareddoran doran says:

    u have helped me a lot in the past two days I wish I could just skip class and watch u instead because my professor is useless compared to ur videos . finding u may have made it possible for me to still pass thank u so very much. keep posting videos and ill keep watching thank u thank u thanku

  • arman sesar says:

    i hear you clearing your throat too often, that chuck cant be good for your lungs, maybe its time for a white board?

  • Hanson Liao says:

    The 9 lol

  • Ali Husain says:

    I know it's been over two years since this video was made, but I just wanted to say thanks for finally helping me understand multiple angles. I'll definitely be checking back to this channel whenever I need help in mathematics.

  • Mohammed Alyafey says:

    Thank you so much for helping and saving my life. You are much better than my teacher. I will make sure to tell everyone about your channel. Btw, do you have a video about sketching trigonomentric functions?

  • L Xiong says:

    You are my Spirit Animal… Thank you so much for making this video! Cried tears of relief

  • Josefina Buzzard says:

    How would you solve sin x + 4 sec x + 5 = 0?

  • Thuy Nguyen says:

    Wow, that shirt looks great!  Math is awesome when taught handsomely.

  • Matt Bingham says:

    This dude is really annoying me…

  • Bongumenzi Freeman says:

    This guy when he appears in his most videos he like to jump  ….like wow that cool entertainment while we learn super dope though

  • Scieneering says:

    if the problem is say 2cos(3x)=1 after we get cos by itself as cos(3x)=1/2 do we then multiply both sides by 1/3? to get rid of the 3 like you were doing with 1/2 and 2?

  • Lindsey Turner says:

    Thank you. I have been stuck for two days with no help. You have saved my life.

  • bubba1119054 says:

    I'm positive I would not be passing math class without your videos… thank you sir.

  • Jenna Barzak says:

    I have to say that you are absolutely fantastic. I'm taking a trig final today and I have learned more in the last few days studying from your videos than I have this entire semester. Thank you for being so clear and easily understandable.

  • Chloe Jackson says:

    Hiya 🙂 I don't quite understand what you say when you mean go all the way around the circle. For example, in the example problem one, I see where you pulled the (eventual) pi/6 and 5pi/6 from, but where did the 7pi/6 and 11pi/6 come from? Any insight would be appreciated. Love love LOVE your videos by the way. You're a life saver!

  • Jasmine Gault says:

    I know this video is older but I'm currently learning this and this was super helpful! The only thing different is my teacher wants us to use the format of answers + 2 pi k, where k is an integer. It's a lot more confusing for me to figure out how to write those intervals than to just go around the unit circle and list all the rest of the possible answers.

  • Smit Modi says:

    How did u get the values for 2x= pi/3 and 5pi/3 (2:25 to 2:35)

  • ar.1998 says:

    Why is the interval opened at 2 pi ?? Why it is not included here while in video 33 in the playlist where you where talking about single angles limited solutions the restriction interval was closed and 2pi was included ??

  • Neal Angelo Lazado says:


  • Joyzi Ami says:


  • Novakira Sato says:

    This guy belongs in the 1950's…

    Also, how did he get another 1/2 after taking the arccos of 1/2? I don't understand why he's multiplying anything by 1/2 again since π/3 [ quadrant 1] and 5π/3 [ quadrant 4 ] are the only possible results for a positive arccos (1/2) within a 2π period.

  • Emma Peneguy says:

    Thumbs up for singing salt n pepa <3

  • Emma Peneguy says:

    ok so I have 2cos^2(x)+3cos(x)+1=0 …somebody help

  • Sam Cantaloup says:

    I love your videos! Soso helpful! THANKYOU!

  • Flyer Aerospace says:

    Mr. True… you are a GOD-SENT for all students around the world. I would recommend your videos to all my classmates. Please, keep the videos rolling and hopefully with enough subscriptions you would be able to quit your day job and start teaching math to the entire world. THANK YOU VERY VERY MUCH!

  • Tyler Snick says:

    good job explaining but your speed is really fast for how mathematically slow and stupid I am, thnx for the vid.

  • Darkwing Dumpling says:

    I thought the domain and range of inverse functions were restricted, like arccos range is from 0 to π so wouldn't 5π/3 be invalid answer? That's what my instructor tells me. Why it seems like sometimes you're allowed to cross that range and other times it is invalid?

  • Jason Carbon says:

    Thanks for the help.

  • Michael Liang says:

    Can you explain if it matters if you applied the transformation (horizontal compression/stretch or whatever) before you found the other angles in different quadrants? Like that first ex.: you found 2x in the different quadrants then isolated for x. Would it be wrong to isolate x first before finding cosine in the different quadrants? I've heard the phrase, "It will change the period (dummy)" but I still don't get it T__T.


  • Tinpoppy says:

    Thank you so much!!

  • Najeeb Anwar Tareen says:

    you have beaten khan academy, your way of explaining is awesome love the way how you keep balance between entertainment and study you are good at getting our focus and BAM!!!!!!!!! gotta do my homework :)that's the end. keep smiling

  • Ariel Haber-Fawcett says:


  • Marcia Amerson says:

    Great video as always. I have been learning so much from your videos. I have a question @ProfRobBob. What if I had sinx 2x + sin 5 x. How would I approach this problem? And ifi HAD SINX 2X + sin 6X . How would you set it up? Would it be 2sin 2X.cos2x. Thank you

  • Jasmine Falls says:

    You're really smart, but I feel as if you were forgetting simple math whenever you would glance over at whatever it was that you were looking at. Just making an observation, nevertheless, great review thanks.

  • Joyce Tng says:

    I LOVE YOU!!!!!!!!

  • chris b says:

    Thanks for your help. You really nailed it! Super appreciative of your valuable time spent helping us math students. Take care !

  • David marsden says:

    how come you don't use the identities to solve these?

  • Sarku Roll says:

    Super helpful

  • Olawunmi Akisanya says:

    Honestly, you're an AMAZING Teacher. you make maths so much easier and your enthusiasm makes it fun. I've had trouble for days with this topic and after this video, it finally makes sense. Thank you so much.

  • Luis torres says:


  • Abdylkader Cerkezi says:

    arcos(sinx) can you slove ?

  • A E says:

    cool correction!

  • John Doe says:

    I wish my math teacher explained it this good…

  • Amelia Morrison says:

    Man I love this guy, I am so relieved by him explaining trig so understandably, I sometimes laugh from the shock of how easy it ends up being, as opposed to how my book and professor make this stuff look. THANK YOU! !

  • boozapian says:

    From someone brushing up on their math skills in order to become certified to teach math, I LOVE your videos. This one was particularly helpful. I had been incredibly frustrated prior to watching this. Just wanted to let you know, it's not just young students watching your videos!

  • Pang Layman says:

    If it is of any help: To answer the question (using Mr Tarrou's first example) of why we get what "appear" to be negative "answers" for the evaluation of cos-1(1/2). In the UNIT CIRCLE our answers for the function cos-1(1/2) would be pi/3 and 5pi/3, – ONLY – that is correct thinking.

    HOWEVER, what we are trying to do here is find all angle values that get "cut back" (divided by the coefficient in front of x) to fall within the unit circle (that is, between 0 and 2pi).

    In the first example: The function cos-1(1/2) yielded: pi/3 and 5pi/3 (both positive x-axis values). These then get cut back (by 1/2) to give pi/6 and 5pi/6 (a negative x-axis value).

    Because we are cutting back, this means that there "could be" some other angles out there (greater than 2pi), that when cut back by (1/2) also fall in the unit circle. Mr Tarrou then produced 2 more angle values: 7pi/3 and 11pi/3 (both positive x-axis values). These then get cut back to ( by 1/2) to yield: 7pi/6 (a negative x-axis value) and 11pi/6.

    So the answer contains two angles 5pi/6 and 7pi/6 that "appear" wrong at first sight, but really are not.

  • bablysnukly2 says:

    Hands down one of the best math teachers, ever. Thank you so much.

  • Ujjwal Mehrotra says:

    that jump lol

  • Jagan Argos says:

    Since the 2pi wasn't included and 0 could be, would you list it as 0 instead or just remove it from answers?

  • Austin In says:

    Man I wish you were my real Trig Professor! I've watched so many of your videos since my second exam was not to good (66%) my third exam is next week, lets see if it works! Thank you for all your help!

  • Andrew says:

    You are awesome man! I was struggling with these so much ty.

  • Toe B says:

    Cos (5pi/6) does not equal 1/2 it should be (4pi/6) and (10pi/6) on your first example

  • Pranesh Balasubramaniam says:

    Haha I love the 9

  • Vanessa Flores says:

    You were my savior during high school. I watched your videos while I would be doing my homework and you saved my grades! I stopped watching your videos in college for a bit until today when my teacher put this video under her "helpful websites" tab. When I saw your name, I was like "oh, okay this person sounds familiar." AND THEN, when you jumped on the screen and I flipped out and yelled YYYAAASSSS!!! My mom even asked if I was going crazy. HAHA! I am forever grateful to you. I hope you keep teaching!! I absolutely love you! THANK YOU!!!!!

  • Kenny Zelaya says:

    i saw how you caught that eraser and great video you're very clear with your videos thanks. 👍👍👍

  • yabg douglas says:

    sir you have saved my life before a test bam! thank you so so much!

  • christiana lauridsen says:

    how tf was he getting those radians from the unit circel i dont get that at all

  • giger h.r says:

    My math professor is not good at teaching . You are best Professor.

  • Patrick Curran says:

    This video was so helpful and fun!

  • Musab Ali says:

    How did u get 5pi over three, ProfRobBob?

  • SNoCappidona says:

    pause at 11:31

  • Cyrus says:

    Mr. True…I have been stuck dealing with this type of equations. Thanks a lot!

  • Adam Butts says:

    Can you do a demonstration with decimals

  • Michael Courter says:

    These videos are a couple of years old, but you are an amazing teacher. Thank you for posting all of your videos.

  • Adam Seidman says:

    I really appreciate all the videos. I like your technique on explaining multiple angles in a trig equation. My question is on Question #3. If you graph the function, x=0. This value was not found algebraically. If you check the function with x=0, it works. Just not sure how to account for the zero in the computation besides finding it in the graph.

  • Cheesy Par says:

    My teacher never mentioned how the coefficient is how many times you go around the unit circle (if he did, I wasn't paying attention), and that really helps. Now I understand this way more! Thank you!

  • jenna says:

    cannot thank you enough!!!!!!!

  • SuperPaulBF says:

    Thank you so much for this video and all that you do. I dropped out of high school when I was sixteen and didn't start college until I was 23. When I went back to school, I had to start in the lowest remedial math course… It was embarrassing and difficult but one of my friends who was a math wiz told me about your videos and ever since I've been a huge fan. I've worked my way up through all the remedial courses and college algebra, all with straight A's, and I'm just about to finish pre-calculus with an A as well. I say all of that because I seriously don't think I could have done it without your help. I've had amazing math teachers through college so far, but you're still the best by far. I literally tell everyone on my campus about this channel. Once again, thank you so much. And now it's time to go do my homework! 🙂

  • advisualz says:

    you the goat

  • Nikhil Shanbhag says:

    extremely helpful video. keep it up!

  • JayLink11 says:

    That handwriting is beautiful…

  • Ranunculus says:

    You probably won't see this because it's been so long, but I just want to thank you for uploading this video and putting in the effort to save other people's headache and grades. I was recently dealing with an equation with 4 theta, and I looked for help for hours. And then I discovered you, who taught me how to do it in less than 30 minutes. Thank you, you earned a sub.

  • soso Liwa says:

    Realllly thank you you have cleared thousands of questions
    But there is a question that I asked for no one answered me pleaaaase explain it for me,you know one question would make the whole chapter not understood,
    The question is and stay with me step by step ,didn’t we say that there is restriction for the inverse of cos and son and tan ,for example the restriction of the inverse of cos is 0to 180.Right…
    So why we didn’t restrict that here to it is an inverse so why not
    Pleeeeaaaaase answer me
    And I am so thankful to you

  • Meta Finals says:

    Thank you so much man. Mymathlab can never go into detail the way you did.

  • Aaron Hall says:

    x also equals zero for the second to last example.

  • Vikram karmarkar says:

    do you have any video on this same topic where you do an example that has both sine and cosine in the same equation and you have to solve that? I am really struggling with those. Thank you for your help!

  • Brenna Osborne says:

    Thank you so much! I have been stuck on these for so long and I watched 10 minutes of your video and I'm getting all of these questions right!

  • Nonster87 says:

    This video helped me a lot. Taking precalculus online was my only option so ive been watching a lot of your videos.

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