Question

How do you solve `cos^2 x = sin^2 (x/2)` ?

Answer 1

`x = +-pi/3 + 2npi" "` or `" "x = (2n+1)pi" "` for any integer `n`

Explanation:

Note that:

`cos(2 theta) = cos^2 theta - sin^2 theta`

`color(white)(cos(2 theta)) = 1-2sin^2 theta`

So:

`2 sin^2 theta = 1 - cos(2 theta)`

Putting `theta = x/2`, we find:

`2 cos^2 x = 2 sin^2(x/2) = 1 - cos x`

Adding `cos x - 1` to both ends, this becomes:

`0 = 2 cos^2 x + cos x - 1 = (2 cos x - 1)(cos x + 1)`

So:

`cos x = 1/2" "` or `" "cos x = -1`

If `cos x = 1/2` then `x = +-pi/3 + 2npi" "` for any integer `n`

If `cos x = -1` then `x = (2n+1)pi" "` for any integer `n`