Trigonometric equation, find $\sin \theta $ Find $\sin \theta $ if $a$ and $c$ are constants
$$ 1-\left(c-a\tan\theta\right)^2=\frac{\sin^2\theta\cos^4\theta }{a^2-\cos^4\theta } $$
 A: I think this relust is ugly.let
$$t=\sin{x}$$
$$1-\left(c-a\cdot\dfrac{t}{\pm \sqrt{1-t^2}}\right)^2=\dfrac{t^2(1-t^2)^2}{a^2-(1-t^2)^2}$$
so
$$(1-t^2)-(\pm c\sqrt{1-t^2}-at)^2=\dfrac{t^2(1-t^2)^3}{a^2-(1-t^2)^2}$$
then
$$[a^2-(1-t^2)^2][(1-t^2)-(c^2(1-t^2)-\pm 2act\sqrt{1-t^2}]=t^2(1-t^2)^3$$
A: after simpßlification we find this here
$$2\,\sin \left( \theta \right)  \left( \cos \left( \theta \right) 
 \right) ^{5}ac- \left( \cos \left( \theta \right)  \right) ^{8}+{a}^{
2} \left( \cos \left( \theta \right)  \right) ^{6}- \left( \cos
 \left( \theta \right)  \right) ^{6}{c}^{2}+2\, \left( \cos \left( 
\theta \right)  \right) ^{6}- \left( \cos \left( \theta \right) 
 \right) ^{4}{a}^{2}-2\,\sin \left( \theta \right) {a}^{2}c\cos
 \left( \theta \right) -{a}^{3} \left( \cos \left( \theta \right) 
 \right) ^{2}+a{c}^{2} \left( \cos \left( \theta \right)  \right) ^{2}
-a \left( \cos \left( \theta \right)  \right) ^{2}+{a}^{3}
=0$$
i think it is implossible to find an explicit formula for $\theta$ try a numerical method
