Question

Solve 2-sin^2x = 2cos^2(x/2) for both x [0,2pi) and all real x.

Answer 1

Do you mean \(2 - \sin^2 x = 2 \cos^2(\frac{x}{2})\)?

Answer 2

yes, my bad!

Answer 3

No, you did nothing wrong. I was just clarifying.
What have you tried? What do you think you can do?

Answer 4

I felt like it would come down to a variation on a pythagorean identity, like cos^2x=1-sin^2x= (1-sinx)(1+sinx) but I don't think it's possible to get there with cos^2(x/2). Would it be easiest to use power reducing and half identities?

Answer 5

Yes. Try that.