distinct roots of the equation $A\sin^3 x+B\cos^3 x+C =0$ 
The number of distinct roots of the equation $A\sin^3 x+B\cos^3 x+C =0$
no two of which differ by $2\pi$ is, Where $A,B,C\in \mathbb{R}$

$\bf{(a)}\;\;\;\;\;\; 3\;\;\;\;\;\; (b)\;\;\;\;\;\; 4\;\;\;\;\;\; (c)\;\;\;\;\;\; 6\;\;\;\;\;\; (d)\;\;\;\;\;\; infinite$
$\bf{My\; Try::}$ Using $$\sin 3x =3\sin x -4\sin^3 x\Rightarrow 4\sin^3 x=3\sin x-\sin 3x$$ and
$$\cos 3x = 4\cos^3 x-3\cos x\Rightarrow 4\cos^3 x=\cos 3x+3\cos x$$
So we get $$A(4\sin^3 x) +B(4\cos^3 x)+4C=0$$
So $$A(3\sin x-\sin 3x)+B(\cos 3x+3\cos x)+4C=0$$
I did not understand How can I solve that question, Help me
Thanks

Here, I found a restriction for the constants $A$, $B$, $C$ that wanted to share. We know by the well-known inequalities $|\sin x| \le 1$ and $|\cos x| \le 1$ that
$$-(|A|+|B|) \le A\sin^3 x+B\cos^3 x \le (|A|+|B|)$$
So, if we wish the equation to have a root we must have
$$-(|A|+|B|) \le -C \le (|A|+|B|)$$
Or equivalently
$$|C| \le |A|+|B|$$
