Question

$1 - 2 \cos(x) > 0$
$\cos(2x + \frac{\pi}{6} ) < 0.5$
$\cos( \frac{\pi}{4} ) \times \cos(x) - \sin( \frac{\pi}{4} ) \times \sin < - \frac{ \sqrt{3} }{2}$
помогите please​

Answer 1

$1)\; \; 1-2cosx>0\; \; \Rightarrow \; \; \; cosx<\frac{1}{2}\\\\\frac{\pi }{3}+2\pi n<x<\frac{5\pi}{3}+2\pi n\; ,\; n\in Z\\\\\\2)\; \; cos(2x+\frac{\pi }{6})<0,5\\\\\frac{\pi}{3}+2\pi n<2x+\frac{\pi}{6}<\frac{5\pi}{3}+2\pi n,\; n\in Z\\\\\frac{\pi }{6}+2\pi n<2x<\frac{3\pi }{2}+2\pi n\; ,\; n\in Z\\\\\frac{\pi}{12}+\pi n<x<\frac{3\pi }{4}+\pi n\; ,\; n\in Z$

$3)\; \; cos\frac{\pi}{4}\cdot cosx-sin\frac{\pi}{4}\cdot sinx<-\frac{\sqrt3}{2}\\\\cos(x+\frac{\pi}{4})<-\frac{\sqrt3}{2}\\\\\frac{5\pi }{6}+2\pi n<x+\frac{\pi }{4}<\frac{7\pi }{6}+2\pi n\; ,\; n\in Z\\\\\frac{7\pi }{12}+2\pi n<x<\frac{11\pi }{12}+2\pi n\; ,\; n\in Z$

Answer 2

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