Question

Помогите на фото......

Answer 1

Ответ:

$\displaystyle \sqrt{ctgx}\Big(sin^2x-\frac{1}{4}\Big)=0\ \ ,\ \ \ \ ODZ:\ \ ctgx\geq 0\ ,\ sinx\ne 0\ \ ,\\\\a)\ \ \sqrt{ctgx}=0\ \ ,\ \ ctgx=0\ \ ,\ \ x=\frac{\pi }{2}+\pi n\ ,\ n\in Z\\\\b)\ \ sin^2x=\frac{1}{4}\ \ ,\ \ \frac{1-cos2x}{2}=\frac{1}{4}\ \ ,\ \ 1-cosx=\frac{1}{2}\ \ ,\ \ cosx=\frac{1}{2}\ \ ,\\\\2x=\pm \frac{\pi}{3}+2\pi k\ \ ,\ \ x=\pm \frac{\pi}{6}+\pi k\ \ ,\ k\in Z\\\\x=-\frac{\pi}{6}+\pi k\notin ODZ\\\\Otvet:\ \ x=\frac{\pi}{2}+\pi n\ ,\ \ x=\frac{\pi}{6}+\pi k\ \ ,\ n,k\in Z\ .$

$c)\ \ x\in \Big[-\dfrac{3\pi }{2}\ ;\ 0\ \Big]:\ \ x_1=-\dfrac{5\pi }{6}\ ,\ \ x_2=-\dfrac{\pi}{2}\ ,$