Question

Помогите пожалуйста решить неравенство
Sinx - (√3)cosx<-1

Answer 1

$\sin(x) - \sqrt{3} \cos(x) < - 1 \\ \sqrt{ {1}^{2} + ( \sqrt{3}) {}^{2} } \sin(x - \arctg( \frac{ \sqrt{3} }{1} )) < - 1 \\ \sqrt{2} \sin(x - \frac{\pi}{3} ) < - 1 \\ \sin(x - \frac{\pi}{3} ) < - \frac{ \sqrt{2} }{2} \\ x - \frac{\pi}{3} \in ( - \frac{3\pi}{4} + 2\pi n; - \frac{\pi}{4} + 2\pi n) \\ x \in ( - \frac{3\pi}{4} + \frac{\pi}{3} + 2\pi n; - \frac{\pi}{4} + \frac{\pi}{3} + 2\pi n) \\ x \in ( - \frac{5\pi}{12} + 2\pi n; \frac{\pi}{12} + 2\pi n), \: n \in \mathbb Z$