Question

Срочно, пожалуйста
Решите неравенство корень из (1+2cosx) > sinx

Answer 1

Ответ:

$\sqrt{1+2cosx}>sinx\ \ \Leftrightarrow \ \ \ \left[\begin{array}{l}\left\{\begin{array}{l}sinx\geq 0\\1+2cosx>sin^2x\end{array}\right\\\\\left\{\begin{array}{l}sinx<0\\1+2cosx\geq 0\end{array}\right\end{array}\right\\\\\\a)\ \ \left\{\begin{array}{l}sinx\geq 0\\1+2cosx>sin^2x\end{array}\right\ \ \left\{\begin{array}{l}2\pi n\leq x\leq \pi +2\pi n\ ,\ n\in Z\\1-sin^2x+2cosx>0\end{array}\right$

$\left\{\begin{array}{l}2\pi n\leq x\leq \pi +2\pi n\ ,\ n\in Z\\cos^2x+2cosx>0\end{array}\right\ \ \left\{\begin{array}{l}2\pi n\leq x\leq \pi +2\pi n\ ,\ n\in Z\\cosx\, (cosx+2)>0\end{array}\right$

$\left\{\begin{array}{l}2\pi n\leq x\leq \pi +2\pi n\ ,\ n\in Z\\cosx<-2\ \ ili\ \ \ cosx>0\end{array}\right\ \ \left\{\begin{array}{l}2\pi n\leq x\leq \pi +2\pi n\ ,\ n\in Z\\-\dfrac{\pi}{2}+2\pi k<x<\dfrac{\pi}{2}+2\pi k\ ,\ k\in Z\end{array}\right\\\\\\x\in \Big[\ 2\pi n\ ;\ \dfrac{\pi}{2}+2\pi n\ \Big)$

$b)\ \ \left\{\begin{array}{l}sinx<0\\1+2cosx\geq 0\end{array}\right\ \ \left\{\begin{array}{l}-\pi +2\pi n<x<2\pi n\ ,\ n\in Z\\cosx\geq -\dfrac{1}{2}\end{array}\right$

$\left\{\begin{array}{l}-\pi +2\pi n\<x<2\pi n\ ,\ n\in Z\\-\dfrac{2\pi}{3}+2\pi k\leq x\leq \dfrac{2\pi}{3}+2\pi k\ ,\ k\in Z\end{array}\right\ \ \ x\in \Big[-\dfrac{2\pi}{3}+2\pi n\ ;\ 2\pi n\ \Big)$

$Otvet:\ \ x\in \Big[-\dfrac{2\pi }{3}+2\pi n\ ;\ \dfrac{\pi}{2}+2\pi n\ \Big)\ ,\ n\in Z$