Solve the trigonometric equation: $\sin {3x} = 4 \sin^2 x$ Solve the equation $\sin{3x} = 4 \sin^2 x$.

I tried to change the $\sin{3x}$ to $3\sin x\cos x$ then solve it, but I could not find the correct answer.
Hint: We have $$\sin 3x = 3\sin x - 4\sin^3 x\neq 3\sin x\cos x$$

So you get (let $\sin x = \alpha$) $$3\alpha - 4\alpha^3 = 4\alpha^2 \iff 4\alpha^3 + 4\alpha^2 - 3\alpha = 0$$

Which is a simple cubic in $\alpha$. Find the roots of the cubic, back-substitute and find the corresponding values of $x$.
$\sin(3x)=\sin(2x+x)$
 thus we have:
$$\sin(2x)\cos(x)+\cos(2x)\sin(x)=4\sin^{2}(x)$$
$$2\sin(x)\cos^{2}(x)+(\cos^{2}(x)-\sin^{2}(x))\sin(x)=4\sin^{2}(x)$$
$$\sin(x)(1-\sin^{2}(x))+(1-2\sin^{2}x)\sin(x)=4\sin^{2}(x)$$
$$3\sin(x)-4\sin^{3}(x)=\sin^{2}(x)$$
$$3\sin(x)-4\sin^{3}(x)=\sin^{2}(x)$$
$$3\sin(x)-4\sin^{3}(x)-\sin^{2}(x)=0$$
$$\sin(x)(3-4\sin^{2}(x)-\sin^{2}(x))=0$$

Simply assign $\sin x=a$ then you get $a=6/8$ and $-9/8$ and $a=0$
$x=\sin^{-1}(6/8)$ or $\sin^{-1}(-9/8)$ or $sin^{-}(0)$