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

See after simplifying use double angle formula so $(2\sin x+\sin x-\sin 3x)=2\sin x-2\cos 2x \sin x=2 \sin x(1-\cos 2x)$ same goes for other so it becomes $2\cos x(1+\cos 2x)$ thus simplifying further you get $a \sin x(1-\cos 2x)+b\cos x(1+\cos 2x)+2c=0$ now you just need values of $cos,sin$ where $c$ has a considerable limit according to $\sin,\cos$ . If $a$ and $b$ are big too then its simply infinite values.
