An inequality with trigonometric functions

Solve $$\sin^2 x< \cos^2 x$$ over the real domain.
$$\sin^2 x < 1-\sin^2 x \longrightarrow \sin^2 x < \frac12 \longrightarrow |\sin x|< \frac{\sqrt2}{2}$$.
The solution is $$x\in \bigcup_{k\in \mathbb Z} (k\pi-\frac{\pi}4,k\pi+\frac{\pi}4)$$.