Question

Решите неравенство: 2sin^2x+sinx-1 > 0​

Answer 1

Ответ:

$2sin^2x+sinx-1>0\\\\t=sinx\ \ ,\ \ -1\leq t\leq 1\ \ ,\ \ \ 2t^2+t-1<0\ \ ,\ \ t_1=\dfrac{1}{2}\ \ ,\ \ t_2=-1\ ,\\\\2(t-\frac{1}{2})(t+1)>0\ \ ,\ \ \ znaki:\ \ +++(-1)---(\frac{1}{2})+++\\\\t>\dfrac{1}{2}\ \ \ ili\ \ \ t<-1\\\\\left\{\begin{array}{l}sinx< -1\\sinx>\dfrac{1}{2}\end{array}\right\ \ \left\{\begin{array}{l}x\in \varnothing \\\dfrac{\pi}{6}+2\pi k<x<\dfrac{5\pi}{6}+2\pi k\ ,\ k\in Z\end{array}\right$

$Otvet:\ \ x\in \Big(\ \dfrac{\pi}{6}+2\pi k\ ;\ \dfrac{5\pi}{6}+2\pi k\,\Big)\ ,\ k\in Z\ .$