How do I solve this, answer in degrees?

You should have your basic identities with you, and one of them is:

\cos(2A) = \cos^2A - \sin^2A = 2\cos^2A -1 = 1 - 2\sin^2A

Use one of these and you would get something just like a quadratic equation to solve. That would be easy :)

Can you post what you get? :)

A minor correction: cos(2A) = 1 - 2 sin^2(A)

However, here's another way: use sin^2(x) = 1 - cos^2(x), then apply the quadratic equation to solve for cos(x).

Oops, typo :o

And somehow, I read it as sin squared 2x...