34. Solve the inequality: $\sin x>\cos ^{2} x$.
34. Let's move on to the function $\sin x$. The inequality takes the form $\sin ^{2} x+\sin x-$ $-1>0$. This inequality holds under the condition $\frac{\sqrt{5}-1}{2}<\sin x \leqslant 1$ or $\arcsin \frac{\sqrt{5}-1}{2}+2 \pi k<x<\pi-\arcsin \frac{\sqrt{5}-1}{2}+2 \pi k$.