# 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 equation..it 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:)

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.

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

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.

the first one and im confused

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

you sir, are wonderful

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

i love you

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 >.<

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

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.

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

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.

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!

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

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

you have serious sweg

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

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

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!

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?

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

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

The 9 lol

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.

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?

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

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

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

This dude is really annoying me…

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

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?

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

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

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.

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!

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.

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

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 ??

I LOVE YOU MAN! I PASSED MY FINAL EXAM! TY!

SIR I AM LOVING TRIGONOMETRY BECAUSE OF YOU. BAM!

This guy belongs in the 1950's…

Also, how did he get

another1/2 after taking the arccos of 1/2? I don't understand why he's multiplying anything by 1/2 again sinceπ/3[ quadrant 1]and5π/3[ quadrant 4 ]are the only possible results for a positive arccos (1/2) within a 2π period.Thumbs up for singing salt n pepa <3

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

I love your videos! Soso helpful! THANKYOU!

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!

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

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?

Eyyo!

Thanks for the help.

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.

Thanks

Thank you so much!!

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

THANK YOU SO MUCH!!! I WAS CONFUSED ABOUT HOW MY TEACHER ALWAYS GOT DOUBLE THE AMOUNT OF ANSWERS I DID!

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

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.

I LOVE YOU!!!!!!!!

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

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

Super helpful

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.

COOL TEACHER! THANKS FOR AN AWESOME EXPLANATION!!

arcos(sinx) can you slove ?

cool correction!

I wish my math teacher explained it this good…

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! !

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!

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.

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

that jump lol

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

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!

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

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

Haha I love the 9

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!!!!!

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

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

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

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

This video was so helpful and fun!

How did u get 5pi over three, ProfRobBob?

pause at 11:31

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

Can you do a demonstration with decimals

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

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.

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!

cannot thank you enough!!!!!!!

There ARE

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! 🙂

you the goat

extremely helpful video. keep it up!

That handwriting is beautiful…

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.

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

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

x also equals zero for the second to last example.

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!

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!

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