Question

•••Помогите решить•••

Answer 1

$1)Cosx=-1\\\\x=\pi+2\pi n,n\in z$

$2)2Sin^{2}x-Sinx-1=0\\\\Sinx=m;-1\leq m\leq 1\\\\2m^{2}-m-1=0\\\\D=(-1)^{2}-4*2*(-1)=1+8=9=3^{2}\\\\m_{1}=\frac{1-3}{4}=-\frac{1}{2}\\\\m_{2}=\frac{1+3}{4}=1$

$1)Sinx=-\frac{1}{2}\\\\x=(-1)^{n}arcSin(-\frac{1}{2})+\pi n,n\in z\\\\x=(-1)^{n+1}\frac{\pi }{6}+\pi n,n\in z \\\\2)Sinx=1\\\\x=\frac{\pi }{2}+2\pi n,n\in z\\\\Otvet:(-1)^{n+1}\frac{\pi }{6}+\pi n,n\in z;\frac{\pi }{2}+2\pi n,n\in z$

$3)Sin(\frac{x}{2}-\frac{\pi }{6}+1=0\\\\Sin(\frac{x}{2} -\frac{\pi }{6})=-1\\\\\frac{x}{2}-\frac{\pi }{6}= -\frac{\pi }{2}+2\pi n,n\in z\\\\\frac{x}{2}=\frac{\pi }{6}-\frac{\pi }{2}+2\pi n,n\in z\\\\\frac{x}{2}=-\frac{\pi }{3} +2\pi n,n\in z\\\\x=-\frac{2\pi }{3}+4\pi n,n\in z$