Question

решите неравенство, пожалуйста​

Answer 1

$cos2x+3sinx\geq -1\\\\cos^2x-sin^2x+3sinx\geq -1\\\\(1-sin^2x)-sin^2x+3sinx+1\geq 0\\\\1-2sin^2x+3sinx+1\geq 0\\\\2sin^2x-3sinx-2\leq 0\\\\t=sinx\; \; ,\; \; -1\leq t\leq 1\; \; ,\; \; 2t^2-3t-2\leq 0\\\\D=9+16=25\; ,\; t_1=-\frac{1}{2}\; ,\; \; t_2=2\; ,\\\\2(t-2)(t+\frac{1}{2})\leq 0\qquad [-1\, ]+++[-\frac{1}{2}\, ]---[\; 1\; ]---[\, 2\, ]+++\\\\t\in [-\frac{1}{2}\, ;\, 1\; ]\\\\-\frac{1}{2}\leq sinx\leq 1\\\\-\frac{\pi}{6}+2\pi n\leq x\leq \frac{7\pi }{6}+2\pi n\; ,\; n\in Z\\\\Otvet:\; \; x\in [-\frac{\pi}{6}+2\pi n\; ;\; \frac{7\pi }{6}+2\pi n\, ]\; ,\; n\in Z$