Предмет: Алгебра
Автор: vityamath
Вопрос задан 4 года назад

Решите неравенства:

$|cosx|\geq \frac{\sqrt{3} }{2} \\cosx(2sin2x+\sqrt{3} )\ \textless \ 0$
$\frac{\sqrt{3} }{3} \leq ctgx\leq4$

Ответы

Ответ дал: NNNLLL54

Ответ:

$1)\ \ |cosx|\geq \dfrac{\sqrt3}{2}\ \ \ \Rightarrow \ \ \ \left[\begin{array}{l}cosx\geq \dfrac{\sqrt3}{2}\\cos\leq -\dfrac{\sqrt3}{2}\end{array}\right\ \ \left[\begin{array}{l}x\in \Big[-\dfrac{\pi}{6}+2\pi n\ ;\ \dfrac{\pi}{6}+2\pi n\ \Big]\\x\in \Big[\ \dfrac{5\pi}{6}+2\pi n\ ;\ \dfrac{7\pi }{6}+2\pi n\ \Big]\end{array}\right\\\\\\x\in \Big[-\dfrac{\pi}{6}+2\pi n\ ;\ \dfrac{\pi}{6}+2\pi n\ \Big]\ \cup \ \Big[\ \dfrac{5\pi}{6}+2\pi n\ ;\ \dfrac{7\pi }{6}+2\pi n\ \Big]$

$2)\ \ cosx\cdot (2sin2x+\sqrt3)<0\\\\a)\ \ \left\{\begin{array}{l}cosx>0\\sin2x<-\dfrac{\sqrt3}{2}\end{array}\right\ \ \left\{\begin{array}{l}cosx>0\\-\dfrac{2\pi }{3}+2\pi k<2x<-\dfrac{\pi }{3}+2\pi k\end{array}\right\\\\\\\left\{\begin{array}{l}-\dfrac{\pi }{2}+2\pi k<x<\dfrac{\pi}{2}+2\pi k\\-\dfrac{\pi }{3}+\pi k<x<-\dfrac{\pi }{6}+\pi k\end{array}\right\ \ \Rightarrow \ \ x\in \Big(-\dfrac{\pi}{3} +2\pi k\ ;\ -\dfrac{\pi}{6}+2\pi k\ \Big)$

$b)\ \ \left\{\begin{array}{l}cosx<0\\sin2x>-\dfrac{\sqrt3}{2}\end{array}\right\ \ \left\{\begin{array}{l}cosx<0\\-\dfrac{\pi }{3}+2\pi k<2x<\dfrac{4\pi }{3}+2\pi k\end{array}\right\\\\\\\left\{\begin{array}{l}\dfrac{\pi }{2}+2\pi k<x<\dfrac{3\pi}{2}+2\pi k\\-\dfrac{\pi }{6}+\pi k<x<\dfrac{2\pi }{3}+\pi k\end{array}\right\ \ \Rightarrow \ \ x\in \Big(\ \dfrac{5\pi}{6} +2\pi k\ ;\ \dfrac{3\pi}{2}+2\pi k\ \Big)$

$c)\ \ x\in \Big(-\dfrac{\pi}{3} +2\pi k\ ;\ -\dfrac{\pi}{6}+2\pi k\ \Big)\cup \, \Big(\ \dfrac{5\pi}{6} +2\pi k\ ;\ \dfrac{3\pi}{2}+2\pi k\ \Big)\ ,\ k\in Z$