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.
 A: 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 $x$.
A: $\sin(3x)=\sin(2x+x)$
 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 $sin^{-}(0)$
A: Notice, $$\sin 3x=4\sin^2 x$$ $$\implies 3\sin x-4\sin^3x=4\sin^2 x$$ $$\implies 4\sin^3x+4\sin^2 x-3\sin x=0$$$$\implies \sin x(4\sin^2x+4\sin x-3)=0$$
$$\implies \sin x(2\sin x-1)(2\sin x+3)=0$$ As you have not mentioned any condition of the unknown value $x$ hence writing the general solution for $x$ as follows $$\implies \color{blue}{\sin x=0}\iff \color{blue}{x=n\pi} $$  $$\implies \color{blue}{2\sin x-1=0\ \text{or}\ \sin x=\frac{1}{2}\ or \ \sin x=\sin\frac{\pi}{6}}\iff \color{blue}{x=2n\pi+\frac{\pi}{6}} $$
$$or\ \color{blue}{\sin x=\sin\frac{5\pi}{6}}\iff \color{blue}{x=2n\pi+\frac{5\pi}{6}} $$
$$\implies \color{blue}{2\sin x+3=0\ \text{or}\ \sin x=\frac{-3}{2}}\iff \color{blue}{\sin x\neq \frac{-3}{2}} $$  Now, the complete general solution can be written as $$\color{blue}{x=[n\pi]\cup\left[2n\pi+\frac{\pi}{6}\right]\cup\left[2n\pi+\frac{5\pi}{6}\right]}$$
Where $\color{blue}{\text{n is any integer}}$
