<s>[INST] 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
 [/INST] First, observe that if $A=0$, $B=0$, and $C=0$, then there are infinitely many solutions.  This, however is boring, so we'll exclude this case.
Using the parameterization $\sin(x)=\frac{2t}{1+t^2}$ and $\cos(x)=\frac{1-t^2}{1+t^2}$ for $\sin$ and $\cos$, (and clearing fractions) we get the equation
$$
8At^3+B(1-t^2)^3+C(1+t^2)^3=0.
$$
By expanding, this becomes
$$
(C-B)t^6+3(B+C)t^4+8At^3+3(C-B)t^2+(B+C)=0.
$$
We can use Descartes' rule of signs to bound the number of roots.  There are at most $3$ sign changes since some of the coefficients repeat.  Replacing $t$ by negative $t$, shows the same structure, so there are at most $3$ negative roots.
The one case that is missing is when $t=\infty$, in this case, $\sin(x)=0$ and $\cos(x)=-1$, which is a solution only when $B+C=0$.  In this case, the polynomial also has a root at $t=0$ and the equation above simplifies to
$$
2Ct^6+8At^3+6Ct^2=0
$$
The LHS has at most $2$ sign changes (observe that for $t>0$ or $t<0$, one case will have no sign variation).
This shows that when $A$, $B$, and $C$ are not all zero, there are at most $6$ solutions (and also shows the dependence on the values of $A$, $B$, and $C$). 
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