Question

Sin(x)^2 = cos(x/2)^2

Answer 1

First of all, you can use

$\sin^2(x)+\cos^2(x)=1 \iff \sin^2(x)=1-\cos^2(x)$

and the equation becomes

$1-\cos^2(x)=\cos^2\left(\dfrac{x}{2}\right)$

Now, from the known identity

$\cos^2(x)=\dfrac{1+\cos(2x)}{2}$

we can half all the angles and we get

$\cos^2\left(\dfrac{x}{2}\right)=\dfrac{1+\cos(x)}{2}$

So, the equation has become

$1-\cos^2(x)=\dfrac{1+\cos(x)}{2} \iff 2-2\cos^2(x)=1+\cos(x)$

So, everything comes down to solve

$2\cos^2(x)+\cos(x)-1=0$

The associated equation

$2t^2+t-1=0$

has roots

$t=-1,\quad t=\dfrac{1}{2}$

So, we want one of the following

$\cos(x)=-1,\quad \cos(x)=\dfrac{1}{2}$

Solve for the associated angles and you're done