Question

Решите неравенство: sinx×cosx >0

Answer 1

$\sin x\cdot \cos x=0\Rightarrow \left[ \begin{gathered} \sin x=0 \\ \cos x=0 \\ \end{gathered} \right.\Rightarrow \left[ \begin{gathered} x=\pi k \\ x=\frac{\pi}{2}+\pi k \\ \end{gathered} \right., \; k\in \mathbb{Z}$

$\sin x\cdot \cos x>0\Rightarrow x\in \left( \pi k , \frac{\pi}{2}+\pi k\right), \; k\in \mathbb{Z}$