Then general values of $\theta$ in inverse Trigo sum 
If $\displaystyle \theta = \tan^{-1}\bigg(2\tan^2 \theta\bigg)-\frac{1}{2}\sin^{-1}\bigg(\frac{3\sin 2 \theta}{5+4\cos 2 \theta}\bigg).$ Then general values of $\theta$ is 

Try: Let $\alpha =\tan^{-1}\bigg(2\tan^2 \theta\bigg)\Rightarrow \tan \alpha =2 \tan^2 \theta$
and $\displaystyle \beta = \frac{1}{2}\sin^{-1}\bigg(\frac{3\sin 2 \theta}{5+4\cos 2 \theta}\bigg)\Rightarrow \sin (2\beta) = \frac{3\sin 2 \theta}{5+4\cos 2 \theta} = 3\frac{2\tan \theta}{1+\tan^2 \theta}\cdot \frac{1}{5+4\bigg(\frac{1-\tan^2 \theta}{1+\tan^2 \theta}\bigg)}$
So $\displaystyle \sin (2\beta) = \frac{6\tan \theta}{9+\tan^2 \theta}.$
could some help me how to find range of $\theta$. Thanks

Hint:
$$\dfrac{3\sin2t}{5+4\cos2t}=\dfrac{6\tan t}{9+\tan^2t}=\dfrac{2\cdot\dfrac{\tan t}3}{1+\left(\dfrac{\tan t}3\right)^2}=\sin2y$$ where $3\tan y=\tan t$
$$\implies\sin^{-1}\dfrac{3\sin2t}{5+4\cos2t}=\begin{cases}2\tan^{-1}\dfrac{\tan t}3 &\mbox{if }-1\le\dfrac{\tan t}3\le1\\
\pi-2\tan^{-1}\dfrac{\tan t}3 & \mbox{if }\dfrac{\tan t}3>1\\
-\pi-2\tan^{-1}\dfrac{\tan t}3 & \mbox{if }\dfrac{\tan t}3<-1 \end{cases}$$
For $-1\le\dfrac{\tan t}3\le1,$
$t=\tan^{-1}(2\tan^2t)-\tan^{-1}\left(\dfrac{\tan t}3\right)$
$\implies\tan t=\dfrac{2\tan^2t-\dfrac{\tan t}3}{1+2\tan^2t\cdot\dfrac{\tan t}3}$
Clearly, $\tan t=0$ is a solution.
