# Single Angle Trigonometric Equations All Solutions

BAM! Mr. Tarrou. We are going to do 3 examples

of solving trig equations with single angles where you just have a variable of x, not like

a 2x or 3x. That will be in the next video. It is kind of a repeat of a video I have done

before except this time instead of only accepting answers between 0 and 2pi on the unit circle.

Or 0 to 360. That restriction that my textbook keeps on most trig equation problems. We are

going to find all solutions to the equations. We are going to allow our solutions, allow

ourselves, to rotate around this unit circle in both a positive and a negative direction

an infinite number of times, and thus have all possible solutions to this equation. We

are going to do 3 examples and let’s get started. So we have 2 cos^2 x – 1=cosx. Now I have

a squared term and a single degree term, plus I have a constant so we are going to need

to get this equation set equal to 0 and hopefully factor it. Otherwise we are going to have

to use the Quadratic Formula, which will actually be our 3rd example. This one will factor.

So I am going to move my cosx to the left with subtraction. I am going to get 2cos^2

x – cosx – 1=0. And now we are going to look to see if this trinomial, this quadratic

is factorable. If your 1st term is not equal to 1, and these are both prime numbers so

we could probably guess and check our way to the answer fairly quickly, but if your

1st leading coefficient is not equal to 1, you can take the 1st and last constant, or

the coefficients of 2 and -1, multiply them together and you get some kind of product

which on its own doesn’t mean a whole lot. But we are now going to look for factors of

that product, that -2 that add to the middle coefficient. Of -1. Since 2 is prime it is

not going to be very hard to do. -2 * 1=-2 and -2 + 1=-1, that middle term that I have,

that middle coefficient of -1. I can just write it here to make it a little bit clearer.

So what I am going to do is, I am going to use these two values to rewrite my middle

term and factor by grouping. So we have 2 cos^2 x then break up this middle term using

these two numerical values of -2 and 1 so we are going to have -2cosx + 1cosx, writing

those two terms to make the middle term look different, -1=0. Now in these 1st two terms

I am going to factor by grouping. In these 1st two terms they both have a factor of 2

and a factor of cos. I am going to pull that 2cos out and I am going to get 2cosx. 2cos^2

x divided by 2cosx is cosx. Then 2cosx divided by 2cosx is of course 1. These last two terms

we had cosx – 1. Well inside this parentheses I have cosx – 1. I am looking for a matching

set of factors. I still have to show I am pulling something out of those last two terms

and since it is already a perfect match, I am going to factor out a +1. Which is going

to sort of complete the process, but not really change anything. Of course when I divide by

1 nothing is changed. So I get cosx – 1 again. Now I have two big terms separated

by this one + sign. These two terms both have a cosx – 1. So I am going to once again

sort of undistributed, or factor, that cosx – 1 out. We get cosx – 1, and this term

divided by cosx – 1 is 2cosx + 1. These are all equal to 0. Once you get done factoring

you set each factor equal to 0. We are going to just add 1 to both sides. Subtract 1 and

divide by 2. We get cosx=1 and cosx=-1/2. With the unit circle to look at we can probably

just write down the answers, but to add that extra step of clarity and you may be just

learning this for the first time so you can see all the steps, we are going to get the

cosine function away from the x. It is not cosine times x. It is cosine OF x. It is actually

a math function we know how to do that math function with the inverse function. Kind of

like how you undo addition with subtraction. I am going to undo this by doing the inverse

cosine function, cos^-1, to both sides of the equation. And get x=cos^-1 (1) and x

=cos^-1 (-1/2). Now we are going to draw our attention to the unit circle. Where is

cosine equal to 1? Remember cosine of an angle is x/r and sine of an angle is y/r. If we

are on a unit circle, cosine is simply the x coordinate. What points have an x-coordinate

of 1, well that is 0. SO the cos^-1 (1) is 0. Where on the unit circle is the cosine

value, the x- value equal to -1/2? Well that is here at

2pi/3 and at 4pi/3. These would be my 3 answers if I was looking for my solutions within one

rotation. If I had a restriction that theta had to stay within 0 and 2pi. But I want all

possible solutions. How about I allow myself to keep rotating around the circle. You know

it is a circular function so you are going to have repetition as you continue to rotate

around this 360 degree or 2pi. You will hit these points again. I will hit 1 or 0 degrees

again after I rotate another 2pi. I have degrees on the inside and radians on the outside.

Getting myself confused looking at the inside of the circle, but I have my answers in radians.

I can rotate 2pi and get to 0 again. I can rotate 2pi and get to 0 again. If I am at

2pi/3 I can rotate 2pi and get back to that same coordinate or I can rotate backwards.

The period of sine and cosine is 2pi because you have to go all the way around the circle

to see that number repeat. So I am simply going to say, here is one solution in that

first rotation and I am going to add 2pi*n. I am going to say this because I am running

out of room to write, n is an integer. That means is could be a +3 or -5. If n is allowed

to be an integer, that I’ going to count as do you want to do one full rotation of

2pi to get back to that coordinate here or do you want to do say 3 more rotations of

2pi to get back to that 0, and so on. The n is just going to allow us to say it is an

integer, basically what we are going to put here is the period of the trig function we

are dealing with. This will change more on the 2nd video. So this is going to be 2pi/3

+ 2pi*n, the period of sine or cosine, cosine for this problem adding an integer rotation

to that. The same thing for 4pi/3. That is not just the solutions between 0 and 2pi,

but with that +2pi*n, the period of cosine and n being an integer. I have got all of

our possible solutions. Let’s get this board erased and do our next example. BAM! For our 2nd example we have cot^2 x +

sqrt 3*cotx=0. I have a quadratic again and I already have it set equal to 0, but

my two terms have a common cotangent factor so this factoring is going to be a little

but quicker that my last example. I am going to take this cotx out. cot^2 x divided by

cotx is cotx. sqrt 3*cotx divided by cotx is sqrt3 equals 0. We are done factoring,

let’s set each factor equal to 0. Get the cotangent functions alone so I want to subtract

both sides by sqrt3. Now we have to get the cotangent function away from the x’s. That

is going to be using cot^-1. So x=cot^-1 (0) and x=cot^-1 (-sqrt3). Let’s not forget

the cotangent ration is either cosine/sine or x/y, depending on how your teacher has

been teaching it. Saying x/y might be a little confusing because we have the variable x in

our equation. If we go back to the unit circle the cotangent ratio is x/y and cosine/sine.

Where do I take a cosine and divide it by a sine value or x/y on the unit circle and

get an answer of 0 because it is an inverse trig function so inside that function is that

ratio we have, we are looking for the angle measure. It would be right here x/y or cosine/sin

that is going to be 0/1. When a fraction or a ratio has 0 in the numerator, then the entire

ratio is equal to 0. So the only way for cos/sin which is what cot is to equal 0 is for cosine

to equal 0. So the answers to this are pi/2 and 3pi/2 because you get that x-value of

0 down here as well. As far as my second ratio. I am looking for a cos/sin to equal –sqrt

3 Now cot or tan being negative that must be in quadrant II or IV because the x and

y variables must have different signs. While quadrant I are both positive and quadrant

III are both negative. So we are looking for an answer in quadrant II and IV and I have

a sqrt 3 here but if I do x/y or cos/sin, I am going to have to rationalize that sqrt

3 in the denominator and that will be the right answer we are looking for 5pi/6. Cos

would be –sqrt 3 over 2 divided by ½. Those common denominators would cancel and we would

get a ratio of -sqrt 3. This answer is 5pi/6 and let see I have got the x being sqrt 3

over 2 and the y being -1/2. 11pi/6 would have the same thing. The signs are different,

but we have the same cos and sin value or x and y value on the unit circle. One positive

and one negative, so 11pi/6 is the same answer. Well, it gives you the ratio of –sqrt 3.

Now in the previous example we were using cosine and cosine has a period of 2pi. These

would be the 4 answers if we restricted our answers between 0 and 2pi, but I want all

possible solutions. Why are these being separated? We have that

pi/2 and 3pi/2 are on opposite sides of the circle. I have 5pi/6 and 11/pi/6, those are

on opposite sides of the circle. That is because the period from cotangent is pi and not 2pi.

I could just take one of these answers. I am going to take the smaller one, say if I

want all my solutions I am going to take pi/2 and add it with pi*n. Again pi because it

is the period of cotangent and n could be any integer so we can talk about positive

n’s where we continue the positive rotation or negative n’s where we do the negative

rotation. I am going to pick the smaller of these two values so I am going to pick 5pi/6

and write that right here. Together these are all solutions to this equation. We have

one more example that will involve the Quadratic Formula if you want to take that examples

of trig equations. The

last example is not going to factor. We are

asked to solve 4sin^2 x – sinx – 1=0. I know this is not going to factor because

my leading coefficient is not one. So I am going to take 1st and last, multiply them

together and get a -4. There are no factors of -4 that add to be middle coefficient of

-1. So this will not factor. Thus, we are going to use the Quadratic Formula to finish

this answer and we are not going to get answers that are right off the nice unit circle. We

are going to have some decimal form of these answers. That might make it a bit difficult

to understand or identify all the possible answers, so I am going to help you with that.

a=4, b=-1, c=-1 and we are going to use the Quadratic Formula, which is x equals

opposite of b plus or minus the square root of b squared minus 4ac all over 2a. It is

x equals right, but my variable is not x, it is sin x. So, I am going to have the sin

x equals the opposite of b plus or minus the square root of b squared minus 4ac all over

2a. I am going to pause the video and finish writing this out to shorten this up just a

little bit. You may want to try and solve this yourself and then see if my solution

matches yours, and be back in just a second. I plugged in my values of b, b, a, and c,

and a. -1^2=1, 4*4=16 and two negatives make a positive. My quadratic comes out to

be 1 plus or minus the sqrt 17 all over 8, which simplifies down to 0.64 and -0.39. So,

sin x=0.64 and sin x=-0.39. I apply the sin^-1 function on both of these equations

and I get 39.8 degrees, I am going to do degrees because once you start getting decimal answers

of angles I think it make it makes it a little bit easier, and -23 degrees. Now, let’s

talk about this 39 degrees. Sine is positive, the sin of an angle x gives you 0.64. Sin

is not positive in only one quadrant. Sine being y/r is positive in two quadrants while

the calculator is able to give me an answer in Quadrant I, I need to find my own answer

in Quadrant II. Which will have this same reference angle of 39.8 degrees. So I am going

to go over here to 180 and back up, because I want to stay in Quadrant II and go up, or

subtract 39.8 degrees. So 180 – 39.8=140.2. And draw my attention over here, just out

of habit, because I don’t like getting negative answers for angle measures. The calculator

was only able to give me -23 degrees. Well, the sin x is equal to a negative number, that

y/r is a negative ratio, it means that I must be getting sine values that are in Quadrant

III and IV. So I want an angle in Quadrat III that has a reference angle of 23 degrees.

So I am going to do 180 + 23 and get an angle of 203 degrees, roughly because this is a

bit off. And I want an angle in Quadrant IV that also has a reference angle of 23 degrees.

Remember reference angle is an acute angle and the number of degrees the terminal side

is away from the x-axis. I want something in Quadrant IV because sine is negative in

Quadrant IV. That is 360 degrees. I want to back up 23 degrees and get 360 – 23 which

is 337. Now, I don’t want all answers between 0 and 2pi. I want all possible solutions.

With the period of sine being 2pi, I am just going to come through with all of these and

add, well I have these in degrees, so period of 2pi is a period of 360 degrees, so I am

going to go to all of these answers and just simply add in 360n. That is 3 examples of

solving for all possible solutions. Solving these trig equations when they have one single

angle. Next video will be for multiple angles. I am Mr. Tarrou. BAM! Go do your homework.

nice, I have a test on this subject this week. Thanks man!

You are so welcome…and thank you for being such an awesome viewer!…keep watching and supporting:)

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Mr. Tarrou, y u not my precal teacher??? Dx

We need more teachers like you!! For now though you're one of a kind and a lifesaver of mathematically-challenged teens everywhere!!

Thank you for taking the time to write such an admirable comment:) I'm glad to see you are still watching my videos and I hope you will remain one of my YouTube students throughout the remainder of your math classes.

Oh trust me I will. Pre-cal's already hard enough, I'm going to need all the help I can get for Calculus next year! 😀

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YOUR VIDEOS HELP SO MUCH!!!!! Much bette than my teacher hahaha

Thanks for watching and I hope this makes me your go-to YouTube math teacher now!

The difference of (3pi/2)-(pi/2) is pi, not 2pi and the period of cotangent is pi and not 2pi so my example is correct. pi/2+pi is pi/2+2pi/2 which equals 3pi/2. Another pi radians would give you 5pi/2 which is another angle where cot equals 0. I hope I cleared up any confusion. Thank you again for watching and your support!!!

Can you please tell me how i can find out the value of all x's if i do not have that radian chart with me?

If you want to find the cosine and sine of angles that are not on this unit circle you will need a calculator or table of values…if you are referring to just the unit circle you will need to memorize it or understand how to create the unit circle for yourself. You may want to watch Setting Up the Unit Circle.

Thanks! I got it now! Keep up the good videos! 🙂

Your videos really help me!! MASSIVE THANKS

Wow…you look like you might be my youngest viewer ever…lol

You're welcome, glad they helped you so much and I hope you continue to tune-in and tell others:D

thank you so much! this was a big help!

You're welcome…thanks for choosing Tarrou's Chalk Talk to learn from!

Why did i not find this video before my test???? 🙁

I scored an A on my pre cal test special thanks to you, my father and my older brother

Mr Tarroy should be in the Mathematics hall of fame

This is EXACTLY what I was looking for. Thanks so much! Finally some real explanation to these little math laws I was in the dark about. Game Changing!

I thought you were singing the quadratic formula. I had an assignment where I had to sing it. It was basically a parody of pop goes the weasel or something like that. I took the F instead since I had so many A's but due to the amount of hearing that song the formula got stuck in my head. :/

Thank you Mr. T…….things are starting to clear up–JJ

what is your name??

please solve this :Find no of distint solutions of sin5x.cos3x=sin9x.cos7x [x ϵ o,π/2]

Thanks so much Mr.T 🙂

Thank you so much. Your teaching is very clear and detailed, and there's no way I'd understand all of this without your killer videos!

I don't think I can say much more than has already been said, thank you for sharing these videos. They are a lifesaver!

you look better with shirt not tucked in. keep it like that, good info regardless!!! keep vids coming.

I love this guy!!!

Thank you for your videos RobBob! I have difficulty learning in a classroom but i can replay these videos and do it until i get it right. Also, your shirt selections are top notch!

Thank you for the incredibly helpful videos. I can tell that you are very passionate about teaching and I also thank you for that. It is very difficult to find a math teacher who is even remotely passionate about their job at my college. I am preparing for my precalculus final and your videos have decreased my stress and added much clarity. I have subscribed and plan on getting a jump start on Calculus from your channel.

Prof rob. how would i solve all the equation for tan^2x = 2 expressing it in radians and k as an integer. Thank you.

Thanks for the video, the last example was a really big help!

at 4:05 i'm having a hard time understanding what is going on there

btw at 5:25 the inverse cosine doesn't do anything to the values, it's just being technical right?

This was super helpful! I might just pass my math test tomorrow! (fingers crossed)

"Bang! Go do your homework…"

… okay :/

Sir i need some help. how do i know the cot^-1 value in calculator?

Im a little confused with an assignment my teacher gave us last week, heres an example: 6=8cos^2(x+1) Im dividing out the 8, which gives me 3/4= cos^2(x+1); square root both sides giving root3/2=cos(x+1); inverse both sides cos^-1(root3/2)=x+1; at this point inverse cos of root3/2 gives me pi/6, 11pi/6 …………. Looks like x=pi/6-1(shift 1 left units?) +(2pi)n , 11pi/6-1(shift left)+(2pi)n……………this may look like a mess lol my bad

She is asking for the principle root and then the full general solution.

Wow, that just cleared this section. I think I'm gonna pass the test-out for this. Thanks for the help Mr. T!

Amazing!

EXCELLENT TEACHER

You are so close to having 100,000 subscribers! Unless YouTube hasn't updated the public counter yet and you already have. Congratulations! You deserve it! 🙂

Thanks for the last minute study help!

what's the difference between adding 2piN versus piN?

Question doesn't the inverse limit its trig function? If it does, does that mean that certain solutions do not count?

A.o.a ..

what is the full solution of this question with representation of diagram ..

lies in [ 0,2pi]

1) Sin = – 60 .. kindly post it soon.

are there ways of solving trig equations without the use of a unit circle? and what do you mean by period?

Math sucks ass

Thanks Mr. Tarrou, I was very stuck on this

thank you

11:15 – why is 3pi/2 considered an answer? I thought that the range of the cotangent inverse function is restricted to the interval (0, pi), making pi/2 the only possible answer for arccot(0) = x.