Question

Логарифмическое неравенство.
$\sf 2\log_2^2(\sin x)+3\log _2(\sin x)-2\geq 0$

Answer 1

$2log_2^2(sinx)+3log_2(sinx)-2\geq 0\; \; ,\; \; \; ODZ:\; sinx>0\\\\t=log+2(sinx)\; \; ,\; \; 2t^2+3-2\geq 0\; ,\; \; t_1=-2\; ,\; t_2=\frac{1}{2}\\\\2(t+2)(t-\frac{1}{2})\geq 0\; \; ,\; \; \; +++[-2\, ]---[\frac{1}{2}\, ]+++\\\\t\in (-\infty ,-2\, ]\cup [\, \frac{1}{2},+\infty )\\\\\left [ {{log_2(sinx)\leq -2} \atop {log_2(sinx)\geq \frac{1}{2}}} \right. \; \left [ {{sinx\leq 2^{-2}} \atop {sinx\geq \sqrt2}} \right. \; \left [ {{sinx\leq \frac{1}{4}} \atop {x\in \varnothing }} \right. \; \; \to\; \; \; 0<sinx\leq \frac{1}{4}$

$x\in (2\pi n\, ;\, arcsin\frac{1}{4}+2\pi n\, ]\cup [\, \pi -arcsin\frac{1}{4}+2\pi n\, ;\, \pi +2\pi n\, )$