Question

Розвяжить неривнисть

1) cos⁴x-sin⁴x>-✓2/2

2) 2sin²x+✓3sinx≤0

Answer 1

Ответ:

$1)\ \ cos^4x-sin^4x>-\dfrac{\sqrt2}{2}\\\\(cos^2x-sin^2x)(\underbrace {cos^2x+sin^2x}_{1})>-\dfrac{\sqrt2}{2}\\\\cos2x>-\dfrac{\sqrt2}{2}\\\\\dfrac{3\pi}{4}+2\pi n<2x<\dfrac{5\pi }{4}+2\pi n\ \ ,\ \ \ \dfrac{3\pi }{8}+\pi n<x<\dfrac{5\pi}{8}+\pi n\ ,\ n\in Z\\\\x\in \Big(\ \dfrac{3\pi }{8}+\pi n\ ;\ \dfrac{5\pi}{8}+\pi n\ \Big)\ ,\ n\in Z$

$2)\ \ 2sin^2x+\sqrt3sinx\leq 0\\\\sinx(2sinx+\sqrt3)\leq 0\\\\a)\ \left\{\begin{array}{l}sinx\leq 0\\sinx\geq -\dfrac{\sqrt3}{2}\end{array}\right\ \ \ ili\ \ \ \ \ b)\ \left\{\begin{array}{l}sinx\geq 0\\sinx\leq -\dfrac{\sqrt3}{2}\end{array}\right$

$a)\ \left\{\begin{array}{l}x\in [-\pi +2\pi n;2\pi n\ ]\ ,\ n\in Z\\x\in [-\dfrac{\pi}{3}+2\pi k\, ;\, \dfrac{4\pi }{3}+2\pi k\, ],\ k\in Z \end{array}\right\ \ \ b)\ \left\{\begin{array}{l}x\in [\ 2\pi n;\pi +2\pi n\ ]\ ,\ n\in Z\\x\in [-\dfrac{2\pi}{3}+2\pi k\, ;\, -\dfrac{\pi }{3}+2\pi k\, ]\end{array}\right$

$a)\ x\in [-\pi +2\pi n\ ;\ -\dfrac{2\pi}{3}+2\pi n\ ]\cup [-\dfrac{\pi}{3}+2\pi n\ ;\ 2\pi n\ ]\ ,\ n\in Z\\\\b)\ \ x\in \varnothing \\\\Otvet:\ x\in [-\pi +2\pi n\ ;\ -\dfrac{2\pi}{3}+2\pi n\ ]\cup [-\dfrac{\pi}{3}+2\pi n\ ;\ 2\pi n\ ]\ ,\ n\in Z\ .$