Question

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

Answer 1

Ответ:

решение на фотографии.

Answer 2

Ответ:

$2sin^2x+sinx-1<0\\\\t=sinx\ \ ,\ \ -1\leq t\leq 1\ \ ,\ \ \ 2t^2+t-1<0\ \ ,\ \ t_1=\dfrac{1}{2}\ \ ,\ \ t_2=-1\ ,\\\\2(t-\frac{1}{2})(t+1)<0\ \ ,\ \ \ znaki:\ \ +++(-1)---(\frac{1}{2})+++\\\\-1<t<\dfrac{1}{2}\\\\\left\{\begin{array}{l}sinx> -1\\sinx<\dfrac{1}{2}\end{array}\right\ \ \left\{\begin{array}{l}x\ne -\dfrac{\pi}{2}+2\pi n\ ,\ n\in Z\\-\dfrac{7\pi}{6}+2\pi k<x<\dfrac{\pi}{6}+2\pi k\ ,\ k\in Z\end{array}\right$

$Otvet:\ \ x\in \Big(-\dfrac{7\pi}{6}+2\pi k\ ;\ -\dfrac{\pi}{2}+2\pi k\ \Big)\cup \Big(-\dfrac{\pi}{2}+2\pi k\ ;\ ;\ \dfrac{\pi}{6}+2\pi k\ \Big),\ k\in Z\ .$