Графически решить неравенство Квадратный корень из 2 sin x <или= 1

$\sqrt{2\sin x } \le 1 \\ \\ $\left\{ \begin{gathered} 2\sin x \ge 0 \\ 2\sin x \le 1 \\ \end{gathered} \ \ ; \ \ $\left\{ \begin{gathered} \sin x \ge 0 \\ \sin x \le \dfrac{1}{2} \\ \end{gathered} \right.$ \right.$$

$$\left\{ \begin{gathered} x \ge \pi n,n \in Z \\ $\left[ \begin{gathered} x \le \dfrac{\pi}{6} + 2\pi n, n \in Z \\ x \le \dfrac{5\pi}{6} + 2\pi n,n \in Z \\ \end{gathered} \right.$ \end{gathered} \right.$ \ (1)$

$x\in [\pi ; \dfrac{5\pi}{6}] \cup [\dfrac{\pi}{6};0] + 2\pi n, n \in Z$

Ответ: x ∈ [π; 5π/6] ∪ [π/6; 0] + 2πn, n ∈ Z