Question

решите неравенство 2sin3x-1≥ 0

Answer 1

Ответ:

$\frac1{18}(12\pi n+\pi)\leq x\leq \frac1{18}(12\pi n + 5\pi)$

Пошаговое объяснение:

$\displaystyle 2\sin 3x-1\geq 0\\2\sin 3x\geq 1\\\sin 3x\geq \frac12\\\frac16(12\pi n+\pi)\leq 3x\leq \frac16(12\pi n + 5\pi)\\\frac1{18}(12\pi n+\pi)\leq x\leq \frac1{18}(12\pi n + 5\pi)$