Log!

Given equation can be rewritten as: $\left( \log_{\cos x} \sin x \right) \left( \log_{\cos x} \sin x \right) = 4$ $\Rightarrow (\log_{\cos x} \sin x)^2=4$ $\Rightarrow \log_{\cos x} \sin x =\pm2$ $\Rightarrow \sin x = \cos^2 x , \dfrac{1}{\cos^2 x}$ Rejecting the second one since it will give sinx =cosx =1while we know base of log cannot be 1 $\therefore \sin x =\cos^2 x=1-\sin^2 x$ This is a quadratic in sinx which can be easily solved to find $\sin x=\dfrac{-1 + \sqrt 5 }{2}$ Rejecting the negative sign since sinx cannot be less than -1. $x = \arcsin \dfrac{\sqrt 5 - 1}2 + 2k \pi$ $\Large y=\boxed5$