Question

Someone verify my work. Thanks.

Answer 1

For the interval \[0 \le x \le \pi \]

Answer 2

On the first I went with: \[\sin ^{2}(2x) = 2\cos{^2}x\] Do I change the cos into a form of sin and make it all equal to 0?

Answer 3

I think I may have figured it out of someone could verify my work. \[\sin(2x)=\sqrt{2}\cos(x)\]\[2\sin(x)\cos(x)=\sqrt{2}\cos(x)\]\[2\sin(x)\cos(x)-\sqrt{2}\cos(x)=0\]\[\cos(x)[2\sin(x)-\sqrt{2}]=0\]\[\cos(x)=0 , 2\sin(x)-\sqrt{2}=0\]\[\sin(x)=\frac{ \sqrt{2} }{ 2 }\]\[\frac{ \pi }{ 4 } or \frac{ 3\pi }{ 4 }\]