# Thread:

1. ## Finding exact solutions

For a) this is my work:
cos 2θ + cos θ = 0
2 cos^2θ - 1 + cos θ = 0
2cos^2θ + 1 cos θ = 1

1) cos θ = 1
2) 2cos θ + 1 = 1
cos θ = 0

I got those two equations above from factoring out the problem. However The answer in the back of the book says, 60, 180, 300 degrees. I know that what I got for an answer will not come out with those degrees. So can you help find what I did wrong? Thank you so much.

2. Originally Posted by florx

For a) this is my work:
cos 2θ + cos θ = 0
2 cos^2θ - 1 + cos θ = 0
2cos^2θ + 1 cos θ = 1

1) cos θ = 1
2) 2cos θ + 1 = 1
cos θ = 0

I got those two equations above from factoring out the problem. However The answer in the back of the book says, 60, 180, 300 degrees. I know that what I got for an answer will not come out with those degrees. So can you help find what I did wrong? Thank you so much.
Only this part of your work is good:
$cos^{2}\theta + cos\theta = 0$

$2 cos^{2}\theta - 1 + cos\theta = 0$

$2cos^2\theta +cos\theta - 1 = 0$............(1)

you need to find the value of $cos\theta$. For that, suppose $x=cos\theta$ in equation (1) to get:

$2x^2+ x - 1 =0$

Solve this quadratic equation for x, which is $cos\theta$

3. Thank you for another yet excellent and helpful post.

I had figured I made a mistake in typing up my initial problem solving. It was actually:

cos 2θ + cos θ = 0
2 cos^2θ - 1 + cos θ = 0
2cos^2θ + cos θ = 1 (move the 1 over to the right side)
cosθ(2cosθ + 1) = 1 (factor out the left side)

And thus we would have
1) cos θ = 1
2) 2cos θ + 1 = 1
cos θ = 0

But it would have given a wrong answer none the less.

I am wondering how come we leave the 1 on the left side and then factor out the left side and set the right side 0? I thought we would have to move over the 1 to the right side first and then factor out the left side and set it equal to 1...

4. Originally Posted by florx
Thank you for another yet excellent and helpful post.

I had figured I made a mistake in typing up my initial problem solving. It was actually:

cos 2θ + cos θ = 0
2 cos^2θ - 1 + cos θ = 0
2cos^2θ + cos θ = 1 (move the 1 over to the right side)
cosθ(2cosθ + 1) = 1 (factor out the left side)

And thus we would have
1) cos θ = 1
2) 2cos θ + 1 = 1
cos θ = 0

THIS IS WRONG
But it would have given a wrong answer none the less.

I am wondering how come we leave the 1 on the left side and then factor out the left side and set the right side 0? I thought we would have to move over the 1 to the right side first and then factor out the left side and set it equal to 1...
Without moving the 1 to the right side, you have:

$2cos^{2}\theta +cos\theta - 1 = 0$

Let $u = cos\theta$ , which gives you:

$2x^2+x-1=0$

This is a quadratic equation fo the form $ax^2+bx+c=0$ where

$a = 2; b=1; c=-1$

x is given by $\frac{-b \pm \sqrt{b^{2}-4ac}}{2a}$

Now solve for x. Solving for x will give you the value of

$cos\theta$. From there, you can find out what the value(s) of

$\theta$ is(are).

First of all try finding x from the quadratic equation, which will be equal to

$cos\theta$
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