<s>[INST] 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.
 [/INST] 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 $\color{blue}{\text{n is any integer}}$
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