Question

Помогите пожалуйста☺️

Answer 1

$1)cos(x)=-\frac{1}{2}<=>x=\pi+-\frac{\pi}{3}+2\pi k<=>x=\frac{2\pi}{3}+2\pi k,x=\frac{4\pi}{3}+2\pi k\\2)cos(3x)=-\frac{1}{2}<=>3x=\pi+-\frac{\pi}{3}+2\pi k<=>x=\frac{2\pi}{9}+\frac{2\pi k}{3},x=\frac{4\pi}{9}+\frac{2\pi k}{3} \\3)cos(3x-\frac{\pi}{3} )=-\frac{1}{2} <=>3x-\frac{\pi}{3}=\pi+-\frac{\pi}{3}+2\pi k\\3x=\pi+2\pi k=>x=\frac{\pi}{3}+\frac{2\pi k}{3}\\3x=\frac{4\pi}{3} +2\pi k=>x=\frac{5\pi}{9}+\frac{2\pi k}{3}$

$4)cos(\frac{\pi}{3}-3x )=-\frac{1}{2}<=>\frac{\pi}{3}-3x=\pi+-\frac{\pi}{3}+2\pi k\\-3x=\pi+2\pik=>x=-\frac{\pi}{3}+\frac{2\pi k}{3}\\-3x=\frac{\pi}{3}+2\pi k=>x=-\frac{\pi}{9}+\frac{2\pi k}{3}\\5)6cos^2(x)+cos(x)-1=0\\cos(x)=-\frac{1}{2}<=>x=\pi+-\frac{\pi}{3}+2\pi k=>\\=>x=\frac{2\pi}{3}+2\pi k,x=\frac{4\pi}{3}+2\pi k\\cos(x)=\frac{1}{3}<=>x=+-arccos(\frac{1}{3} )+2\pi k$

$6)3sin(a+\pi)+2cos(\frac{3\pi}{2}+\pi )=-3sin(a)+2sin(a)=-sin(a)->0,3$