Solving $a \sin(\alpha) - c \sin^2(\alpha) = b \cos(\alpha) - c \cos^2(\alpha)$ $a, b, c$ are given positive integers. I need $\sin(\alpha)$ or $\cos$ or anything simple with $\alpha$ from the equation:
$$a \sin(\alpha) - c \sin^2(\alpha) = b \cos(\alpha) - c \cos^2(\alpha)$$
 A: With this much information, you can use $$\cos\alpha=\frac{1-\tan^2\frac\alpha2}{1+\tan^2\frac\alpha2}\text{ and } \sin\alpha=\frac{2\tan\frac\alpha2}{1+\tan^2\frac\alpha2} $$ which will give you a Quartic Equation  in $\tan\frac\alpha2$
Once you have solved for  $\tan\frac\alpha2,$ you can easily get $\cos\alpha,\sin\alpha$ using the above formulae.

Alternatively, we can also do the following:
$$b\cos\alpha=c\cos^2\alpha-c\sin^2\alpha+a\sin\alpha=c(1-\sin^2\alpha)-c\sin^2\alpha+a\sin\alpha$$
$$b\cos\alpha=c+a\sin\alpha-2c \sin^2\alpha$$
Squaring we get $$b^2(1-\sin^2\alpha)=(c+a\sin\alpha-2c\sin^2\alpha)^2 $$
On simplification we shall get a Quartic Equation in $s=\sin\alpha$
But unfortunately, the squaring has introduced extraneous roots which need exclusion.
A: In general, lab bhattacharjee's method will work. However, if $a^2+b^2=c^2$, there is a simpler solution.
If $a^2+b^2=c^2$, and $\tan(\theta)=b/a$, then this simplifies to
$$
\begin{align}
0
&=a\sin(\alpha)-b\cos(\alpha)+c\cos^2(\alpha)-c\sin^2(\alpha)\\
&=c\sin(\alpha-\theta)+c\cos(2\alpha)\\
&=c\cos(\pi/2-\alpha+\theta)+c\cos(2\alpha)\\
&=2c\sin\left(\pi/4-\theta/2-\alpha/2\right)\sin\left(\pi/4-\theta/2+3\alpha/2\right)\tag{1}
\end{align}
$$
where we've used $\cos(A)+\cos(B)=2\cos\left(\frac{A+B}{2}\right)\cos\left(\frac{A-B}{2}\right)$.
$(1)$ means either
$$
\pi/4-\theta/2-\alpha/2\equiv0\pmod{\pi}\tag{2a}
$$
or
$$
\pi/4-\theta/2+3\alpha/2\equiv0\pmod{\pi}\tag{2b}
$$
