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.
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...
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