amparvardi wrote: Find all real $x$ in the interval $[0, 2\pi)$ such that \[27 \cdot 3^{3 \sin x} = 9^{\cos^2 x}.\] $\iff$ $3^{3+3\sin x}=3^{2-2\sin^2x}$ $\iff$ $2\sin^2x+3\sin x+1=0$ $\iff$ $\sin x=-1$ or $\sin x=-\frac 12$ Hence the result : $\boxed{x\in\{\frac{7\pi}6,\frac{3\pi}2,\frac{11\pi}6\}}$

amparvardi wrote: Find all real $x$ in the interval $[0, 2\pi)$ such that \[27 \cdot 3^{3 \sin x} = 9^{\cos^2 x}.\] It can be rewritten as $3^{3+3\sin{x}}=3^{2\cos^2{x}}$ which gives $3+3\sin{x} =2\cos^2{x}$. Substitute $\sin{x}=t$ and considering that $\cos^2{x}=1-\sin^2{x}=1-t^2$ we have $3+3t =2t^2$ which gives $t=-1$ and $t=-\frac{1}{2}$ as solutions. From here it's easy to see that $x_1 =\frac{3\pi}{2}, x_2=\frac{7\pi}{6}$ and $x_3 =\frac{11\pi}{6}$. Edit: After I finished writing I realized it's the same solution as pco's. My apologies.