Question

Розвязати нерівність sin5x≥ -√3÷2 ctg 2x÷3≥1

Answer 1

$\sin 5x \ge -\frac{\sqrt3}2\\ - \frac{\pi}{3} \le 5x \le \frac{4 \pi}{3}\\ -\frac{\pi}{15} \le x \le \frac{4\pi}{15}$

$\cot \frac{2x}3 \ge 1\\ \frac{2x}3 \in [0;\frac{\pi}4] \cup [\pi;\frac{5\pi}4] \: \: | \: * \frac32 \\ x \in [0;\frac{3\pi}8] \cup [\frac{3\pi}2;\frac{15\pi}8]$