To solve this trigonometric equation, we can first rewrite it using the identity $\sin^3 x = (1-\cos^2x)\sin x$ and then simplifying.

$\sin^3 x + 3\cos^3 x = 2\cos x$

$(1-\cos^2x)\sin x + 3\cos^3 x = 2\cos x$

Expanding the left side:

$\sin x - \cos^2 x \sin x + 3\cos^3 x = 2\cos x$

Since $\cos^2 x = 1 - \sin^2 x$:

$\sin x - (1 - \sin^2 x) \sin x + 3\cos^3 x = 2\cos x$

$\sin x - \sin x +\sin^3 x + 3\cos^3 x = 2\cos x$

$\sin^3 x + 3\cos^3 x = 2\cos x$

Therefore, the given equation holds true for all values of $x$.