cosx/2=cox in interval[0,2pi)?

Question

I am stumped on this. I got x==0 and 5pi/3 but that wrong. Can anyone give me the correct answers, please? Thanks!

Answer 1

Brackets are your friend.

If you mean cos(x/2) = cos(x), then substitute y=x/2 and use the equality cos(2y) = 2cos^2(y) -1

You have
cos(y) = cos(2y)
= 2cos^2(y) - 1
or 2 cos^2(y) - cos(y) - 1 = 0.

Quadratic equation, solutions are cos(y) = 1 and cos(y) = -1/2.
The first solution yields y = 0, and the second yields y = 2pi/3 or y = 4pi/3.

Since x=2y, we have x=0, x=4pi/3 and x=8pi/3. The latter is greater than 2pi so it won't work. The answer is:
x=0 ----> cos0 = cos0 = 1
x=4pi/3 ---> cos(2pi/3) = cos(4pi/3) = -1/2

Answer 2

Do you mean cos(x)/2 = cos(x)? If you do, that means that cos(x) must be 0. The cosine is a measurement of how far to the right an angle points, so an angle whose cosine is 0 points straight up and down. That means the answers are 90 and 270 degrees, or pi/2 and 3pi/2 radians.