Question

Решите уравнение
√2sinx/3=-1

Answer 1

$\sqrt2\, sin\frac{x}{3}=-1\\\\sin\frac{x}{3}=-\frac{1}{\sqrt2}\\\\\frac{x}{3}=(-1)^{n}\cdot (-\frac{\pi}{4})+\pi n\; ,\; n\in Z\\\\x=(-1)^{n+1}\cdot \frac{3\pi }{4}+3\pi n\; ,\; n\in Z$