<s>[INST] Solve the equation: $\sin 3x=2\cos^3x$ Solve the equation  :
$$\sin 3x=2\cos^3x$$
my try :
$\sin 3x=3\sin x-4\sin^3x$
$\cos^2x=1-\sin^2x$
so:
$$3\sin x-4\sin^3x=2((1-\sin^2x)(\cos x))$$
then ?
 [/INST] Hint:
$$\sin 3x=2\cos^3x$$
$$3\sin x-4\sin^3x=2\cos^3x$$
$$3\sin x(\cos^2x+\sin^2x)-4\sin^3x=2\cos^3x$$
$$\sin^3x-3\sin x\cos^2x+2\cos^3x=0$$
Divide both sides by $\cos^3x$ and $\tan x=t$
$$t^3-3t+2=0$$
$$(t-1)^2(t+2)=0$$
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