Solution: $3^3{\cdot}3^{3sinx}=3^{2\cos^2x}$, hence $2\sin^2x+3sinx+1=0$. $sinx=-1$, $sinx=-\frac{1}{2}$.Hence it is easy to obtain $x=\frac{3\pi }{2}$, $x=\frac{11\pi }{6}$, $x=\frac{7\pi }{6}$.Ok

A problem I could solve! Rewriting this as $3^3 * 3^{3 \sin x} = {(3^2)}^{\cos^2x}$, we arrive to $3^{3 \sin x + 3} = 3^{2 \cos^2 x}$. We can rewrite this as $3 \sin x + 3 = 2 \cos^2 x$. Remembering that $ \cos^2 = 1 - \sin^2 ,$ we can simplify further to get to $2 \sin^2 x + 3 \sin x +1 = 0$. We can reassign variables and solve the quadratic to get the roots: $ \sin x = -1$ and $ \sin x = - \frac{1}{2}$. Since we are looking for solutions on the whole unit circle, we get that the solutions are: $x = \frac{7 \pi}{6}, \frac{11 \pi}{6},$ and $ \frac{3 \pi}{2}$. That's it . EDIT: Sorry for bumping a really old thread/problem... I was excited I actually managed to solve something.