Question

Решить неравенство:

$cos^2x-sin^2x\ \textless \ \frac{\sqrt{2} }{2}$

Answer 1

$cos^2(x) - sin^2(x) < \frac{1}{\sqrt{2}}\\cos(2x) < \frac{1}{\sqrt{2}}\\2x \in (\frac{\pi}{4} + 2\pi n; \frac{7\pi}{4} + 2\pi n)\\x \in (\frac{\pi}{8} + \pi n; \frac{7\pi}{8} + \pi n)$