Question

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tg(п/6-2x)≥1

4cos^2(x/2)>3

Answer 1

$1)\; \; tg(\frac{\pi}{6}-2x)\geq 1\\\\\frac{\pi }{4}+\pi n\leq \frac{\pi }{6}-2x<\frac{\pi}{2}+\pi n\; ,\; n\in Z\\\\\frac{\pi}{4}-\frac{\pi}{6}+\pi n\leq -2x<\frac{\pi }{2}-\frac{\pi}{6}+\pi n\\\\\frac{\pi}{12}+\pi n\leq -2x<\frac{2\pi}{6}+\pi n\\\\-\frac{\pi}{3}+\pi n<2x\leq -\frac{\pi }{12}+\pi n\\\\-\frac{\pi}{6}+\frac{\pi n}{2}<x\leq -\frac{\pi}{24}+\frac{\pi n}{2}\\\\Otvet:\; \; x\in (-\frac{\pi}{6}+\frac{\pi n}{2}\; ;\; -\frac{\pi}{24}+\frac{\pi n}{2}\, ] \;\; ,\; n\in Z .$

$2)\; \; 4cos^2\frac{x}{2}>3\\\\4cos^2\frac{x}{2}-3>0\\\\(2\, cos\frac{x}{2}-\sqrt3)\cdot (2\, cos\frac{x}{2}+\sqrt3)>0\\\\t=cos\frac{x}{2}\; ,\; \; (2t-\sqrt3)\cdot (2t+\sqrt3)>0\; \; ,\; \; +++(-\frac{\sqrt3}{2})---(\frac{\sqrt3}{2})+++\\\\t<-\frac{\sqrt3}{2}\; \; \; ili\; \; \; t>\frac{\sqrt3}{2}\\\\a)\; \; cos\frac{x}{2}<-\frac{\sqrt3}{2}\\\\\frac{5\pi }{6}+2\pi n<\frac{x}{2}<\frac{7\pi }{6}+2\pi n\; ,\; n\in Z\\\\\frac{5\pi }{3}+4\pi n<x<\frac{7\pi }{3}+4\pi n\\\\b)\; \; cos\frac{x}{2}>\frac{\sqrt3}{2}$

$-\frac{\pi }{6}+2\pi n<\frac{x}{2}<\frac{\pi }{6}+2\pi n\; ,\; \; n\in Z\\\\-\frac{\pi }{3}+4\pi n<x<\frac{\pi }{3}+4\pi n\\\\Otvet:\; \; x\in(-\frac{\pi }{3}+4\pi n\; ,\; \frac{\pi }{3}+4\pi n)\cup (\frac{5\pi }{3}+4\pi n\; ;\; \frac{7\pi }{3}+4\pi n)\; ,\; \; n\in Z\; .$