How do i solve this equation ${\mathbb{R}}$: $3 \sin^3x+2 \cos^3x=2 \sin x+\cos x$? How do I solve this equation ${\mathbb{R}}$: 
$3 \sin^3x+2 \cos^3x=2 \sin x+\cos x $?
Note : I have tried using trigonometric transformation but it seems very complicated to get the result .. may there is a clear variable change or some thing as this ...
Thank you for any help .
 A: One possible way is to rewrite the equality as
$$
3\sin^3(x)-2\sin(x)=-2\cos^3(x)+\cos(x).
$$
WARNING: The following step may introduce extra solutions which will need to be checked at the end.
Now, square both sides.
$$
9\sin^6(x)-12\sin^4(x)+4\sin^2(x)=4\cos^6(x)-4\cos^4(x)+\cos^2(x).
$$
Using the trig identity $\cos^2(x)=1-\sin^2(x)$, you can rewrite this entirely in terms of $\sin^2(x)$.  When I did this on scratch paper, I got
$$
13\sin(x)^6-20\sin(x)^4+9\sin(x)^2-1.
$$
Finally, let $y=\sin^2(x)$ and you have the cubic equation 
$$
13y^3-20y^2+9y-1
$$
to solve.
Note: The solutions to this polynomial are not pretty.
A: HINT:
$$\sin x(3\sin^2x-2)=\cos x(1-2\cos^2x)$$
Dividing both sides by $\cos^3x,$
$$\tan x(3\tan^2x-2\sec^2x)=\sec^2x-2$$
$\tan x=u$ $$\implies u\{3u^2-2(1+u^2)\}=1+u^2-2\iff u^3-u^2-2u+1=0$$
A: To solve: $3\sin^{3}x+2\cos^{3}x=2\sin x+\cos x$.
Use the substitution $\sin x = \frac{2\tan \frac{x}{2}}{1+\tan^{2}\frac{x}{2}}=\frac{2t}{1+t^{2}}$
and $\cos x =\frac{1-t^{2}}{1+t^{2}}$.
We become $t^{6}+4t^{5}-7t^{4}-16t^{3}+7t^{2}+4t-1=0$.
Only to solve numerically, 6 solutions, e.g. $t =\tan\frac{x}{2}=0.212475 \rightarrow x = 0.418722$.
This was not a simple problem, maybe there is a mistake in the question.
