Trigonometric equation, find $\sin \theta $

Question

Find $\sin \theta $ if $a$ and $c$ are constants

$$ 1-\left(c-a\tan\theta\right)^2=\frac{\sin^2\theta\cos^4\theta }{a^2-\cos^4\theta } $$

Answer 1

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 $$[a^2-(1-t^2)^2][(1-t^2)-(c^2(1-t^2)-\pm 2act\sqrt{1-t^2}]=t^2(1-t^2)^3$$

Answer 2

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