<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] $$2\cos^3x=3\sin x-4\sin^3x$$
Divide both sides by $\cos^3x$
$$2=3\tan x(1+\tan^2x)-4\tan^3x$$
$$\tan^3x-3\tan x+2=0$$ which is a cubic equation in $\tan x$
Clearly, one of the roots is $\tan x=1$
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