Solution of a trigonometric equation involving double sines,cosines What is the sum of all the solutions of $$\sin^2 (2\sin (x-\frac {\pi}{6}))+\sec^2 (x-\frac {\pi}{2}\tan^2 (x))=1$$ in $[0-4\pi]$ writing $\sec (..)=\frac {1}{\cos (..)} $ .. and rearranging we have $\cos (a).\cos (b)=1$ now as $\cos $ is $-1\leq \cos^2 (..)\leq 1$ . So both a and b have to be $n\pi $ thus for bracket of $\sin (..) $ the solution will be $\frac {\pi}6,\frac {7\pi}{6},2\pi+\frac {\pi}{6},2\pi+\frac {7\pi}{6} $ but I dont know what to do of the next bracket involving $x,\tan (x) $ . Thanks

We have $$\sin^2\left(2\sin\left(x-\dfrac\pi6\right)\right)+\tan^2\left(x-\dfrac\pi2\tan^2x\right)=0$$
For real $x,$
$\sin\left(2\sin\left(x-\dfrac\pi6\right)\right)=\tan\left(x-\dfrac\pi2\tan^2x\right)=0$
$\implies 2\sin\left(x-\dfrac\pi6\right)=m\pi\ \ \ \ (1)$ where $m$ is any integer
Now as $-1\le\sin y\le1$
$-1\le\dfrac{m\pi}2\le1\implies m=0$
$\implies\sin\left(x-\dfrac\pi6\right)=0\implies  x-\dfrac\pi6=r\pi$ where $r$ is any integer
and $x-\dfrac\pi2\tan^2x=n\pi\  \ \ \ (2)$  where $n$ is  any integer
$\implies \dfrac\pi6+r\pi-n\pi=\dfrac\pi2\tan^2\left(\dfrac\pi6+r\pi\right)$
$\iff\dfrac16+r-n=\dfrac16\implies r=n$
