Trigonometric equation, find $\sin \theta$

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

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

• Welcome to MathSE. It is expected that when you pose a question here that you explain your thoughts on the problem and show any work you have done so that we can write a response appropriate to your skill level. Commented Dec 19, 2014 at 10:49

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 $$[a^2-(1-t^2)^2][(1-t^2)-(c^2(1-t^2)-\pm 2act\sqrt{1-t^2}]=t^2(1-t^2)^3$$
• The actual question is Find x if t,B,A are constants $\frac{\left(\left(\frac{t}{B}-\frac{Ax}{\left(1-x^2\right)^{\frac{1}{2}}}\right)^2-1\right)}{x^2-1}=\frac{x^2\left(1-x^2\right)}{A^2-\left(1-x^2\right)^2}$ Commented Dec 19, 2014 at 11:23
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
• simplification do you mean multiplying both sides by $a^2-\cos^4\theta$? Commented Dec 19, 2014 at 13:17