Prove that a trigonometric equation has six distinct roots Show that,in general,the equation $A \sin^3x+B\cos^3x+c=0 $has six distinct roots,no two of which differ by $2\pi$,and that the tangent of their semi-sum is $-\frac{A}{B}$.
My attempt:
I tried to express it as sixth degree equation.
$A \sin^3x+B\cos^3x+C=0 $
$A \sin^3x+B\cos^3x=-C $
Squaring both sides,
$A^2 \sin^6x+B^2\cos^6x+2AB\sin^3x\cos^3x=C^2 $
Does this prove that this equation has six distinct roots,no two of which differ by $2\pi$ and second part i could not prove.Can someone enlighten me in this problem?Thanks in advance..

By setting $u=\tan\frac{x}{2}$, we have to solve:
$$ A\left(\frac{2u}{1+u^2}\right)^3+B\left(\frac{1-u^2}{1+u^2}\right)^3 = -c $$
that is equivalent to finding the roots of the sixth-degree polynomial:
$$ p(u)=(c+B)+3(c-B)u^2+8A u^3+3(c+B)u^4+(c-B)u^6. $$
I do not agree that this polynomial has in general six real roots: by looking at the coefficients of $u^0,u^1,u^2$ we have that Newton's inequality does not hold, so $p(u)$ may have at most $\color{red}{4}$ real roots, since complex roots come in conjugated pairs.
