Question

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tg(п/6-2x)≥1

4cos^2(x/2)>3

Answer 1

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

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

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