Question

Помогите решить б), пожалуйста ​

Answer 1

Ответ:

$\dfrac{sin2x-2cos^2x}{\sqrt{sinx}}=0\\\\ODZ:\ sinx>0\ \ ,\ \ 2\pi n<x<\pi +2\pi n\ ,\ n\in Z\\\\sin2x-2cos^2x=0\ \ ,\ \ \ \ 2\, sinx\cdot cosx-2cos^2x=0\ \ ,\\\\2\, cosx\cdot (sinx-cosx)=0\\\\a)\ \ cosx=0\ \ ,\ \ x=\dfrac{\pi}{2}+\pi k\ ,\ k\in Z\\\\b)\ \ sinx-cosx=0\ |:cosx\ne 0\ \ ,\\\\tgx-1=0\ \ ,\ \ tgx=1\ \ ,\ \ x=\dfrac{\pi}{4}+\pi m\ ,\ m\in Z$

$c)\ \ \left\{\begin{array}{lll}2\pi n<x<\pi +2\pi n\ ,\ n\in Z\\\\\left[\begin{array}{ll}x=\dfrac{\pi}{2}+\pi k\ ,\ k\in Z\\\ x=\dfrac{\pi}{4}+\pi m\ ,\ m\in Z\end{array}\right\end{array}\right\ \ \ \ \Rightarrow \ \ \left[\begin{array}{l}\ \, x=\dfrac{\pi}{2}+2\pi k\ ,\ k\in Z\\x=\dfrac{\pi}{4}+2\pi m\ ,\ m\in Z\end{array}\right\\\\\\d)\ \ x\in \Big[\ \pi \ ;\ \dfrac{5\pi }{2}\ \Big]\ :\ \ x=\dfrac{\pi}{2} +2\pi =\boxed{\dfrac{5\pi }{2}}\ \ ,\ \ x=\dfrac{\pi}{4}+2\pi =\boxed{\dfrac{9\pi }{4}}$