Pre-cal trigonometric equation problem am i correct to factor out a 2 first?
$$2\sin^2 2x =1$$
$$2(\sin^2 x-1) =1$$
$$2\cos^2 x =1$$
$$\cos^2 x ={1\over2}$$
$$\cos x =\pm{\sqrt{2}\over 2 }$$
i'm only looking for solutions from $$0≤ x ≤ 2\pi $$
$$x = {\frac{\pi}{4}},{\frac{7\pi}{4}},{\frac{3\pi}{4}},{\frac{5\pi}{4}}$$
thanks for any corrections

$$
\begin{align*}
&  2\sin^2(2x)=1 \\
\Rightarrow & 2\left( \frac{1-\cos(4x)}{2} \right)=1\\
\Rightarrow&1-\cos(4x)=1\\
\Rightarrow & \cos(4x)=0 \\
\Rightarrow & 4x=\frac{n\pi}{2}, \, n \in \{ 1,3,5,7,\ldots \}\\
\Rightarrow & x=\frac{n\pi}{8}, n \in\{ 1,3,5,7,\ldots \}
\end{align*}
$$
Hence the set is
$$x\in\left\{\frac{\pi}{8},\frac{3\pi}{8},\frac{5\pi}{8},\ldots\right\}$$
