How to solve $\sin^2 (2x) = 2 \sin (2x)$ I am not sure how to solve this:
 $$\sin^2 (2x) = 2 \sin (2x)$$
I thought I could rewrite it like this:
 $$4 \sin^2 (x) \cos^2 (x)  = 4 \sin (x) \cos (x)$$
and maybe like this as well:
 $$ \sin (x) \cos (x)  = 0$$
but I have no idea whether it is correct and how get to the solution ..
thanks for help
 A: HINT:
Let $\sin{(2x)}=t$ and you get a quadratic equation in $t$.
You need to calculate roots of, $t^2-2t=0$.
A: $$\sin^2(2x) = 2\sin(2x)$$
Assign a variable to $\sin(2x)$ to get a simple quadratic equation.
$$t = \sin(2x)$$
Rewrite the equation.
$$t^2 = 2t$$
Now, just solve the equation.
$$t^2-2t = 0 \implies t(t-2) = 0 \implies t = 0 \text{ OR } t = 2$$
Plug in $\sin(2x)$.
For $n \in \mathbb{Z}$, we get the following solutions:
$$t = 0 \implies \sin(2x) = 0 \implies 2x = (\sin^{-1} 0)+2\pi n$$
$$\implies \begin{cases}
2x = 2\pi n \implies \boxed{x = \pi n}\\
\\
2x = \pi+2\pi n\implies \boxed{x = \frac{\pi}{2}+\pi n}\\
\end{cases}$$
We can immediately rule out the second option ($t = 2$) because the range of $\sin x$ is $y \in [-1, 1]$.
Combining the two solutions, we can reach a single general solution: $$\boxed{x = \frac{n\pi}{2}}$$
A: HINT
We have
$$\sin^2 (2x) = 2 \sin (2x)\iff \sin^2 (2x) - 2 \sin (2x)=0 \iff \sin (2x)\cdot [\sin (2x) - 2]=0$$
then recall that
$$A\cdot B=0 \iff A=0 \,\lor\,B=0$$
