Thread: 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.

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:




............(1)

you need to find the value of . For that, suppose in equation (1) to get:



Solve this quadratic equation for x, which is

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...

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:



Let , which gives you:



This is a quadratic equation fo the form where



x is given by


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

. From there, you can find out what the value(s) of

is(are).

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