# Finding exact solutions

• Apr 12th 2010, 09:27 PM
florx
Finding exact solutions
http://i39.tinypic.com/dcg74n.jpg

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.
• Apr 12th 2010, 09:33 PM
harish21
Quote:

Originally Posted by florx
http://i39.tinypic.com/dcg74n.jpg

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 $cos\theta$
• Apr 12th 2010, 10:16 PM
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

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...
• Apr 12th 2010, 11:07 PM
harish21
Quote:

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

$cos\theta$