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 $\cos^2 x = 1 - \sin^2 x$:
$\sin x - (1 - \sin^2 x) \sin x + 3\cos^3 x = 2\cos x$
$\sin x - \sin x +\sin^3 x + 3\cos^3 x = 2\cos x$
Therefore, the given equation holds true for all values of $x$.
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 $x$.