I'm confusing about a question:
For three positive constants $A$, $B$, $C$, find $\theta$ (express in $A$, $B$ and $C$).
$$A\sin^6\theta - A\sin^4\theta + B\sin^2\theta -C = 0$$
So, when I do it the trigonometric way,
$A\sin^4\theta(\sin^2\theta-1) + B\sin^2\theta - C = 0$
$-A\sin^4\theta\cos^2\theta + B\sin^2\theta -C = 0$
$-\frac{A}{4}\sin^2\theta\sin^22\theta +B\sin^2\theta -C = 0$
I found myself unable to continue.
Anyone get some ideas? Appreciate your help.
Taking $x=\sin \theta\implies$ $$Ax^3-Ax^2+Bx-C=0\\\implies x^3-x^2+Dx-E=0,\ D:=\frac{B}{A},\ E:\frac{C}{A}$$ Now use Cardano's method to solve this cubic equation to solve for $x$ and hence finally get $\theta$.