Solving each part separately,
\cos (2x)=0 or 2\cos (x)+1=0
Now,
\cos(2x)=0, 0\le x\le 2\pi
General solutions for cos(2x)=0,
cos(2x)=0 : 2x=pi/2 +2pin , 2x=(3pi)/2+2pin
Solving we get,
2x=\frac{\pi }{2}+2\pi n :x=\frac{4\pi n+\pi }{4}
2x=\frac{3\pi }{2}+2\pi n :x=\frac{4\pi n+3\pi }{4}
x=\frac{4\pi n+\pi }{4},x=\frac{4\pi n+3\pi }{4}
So,solution for the range 0\le x\le 2\pi
x=\frac{\pi }{4},x=\frac{3\pi }{4},x=\frac{5\pi }{4},x=\frac{7\pi }{4}
Again,we have,
2\cos(x)+1=0,0\le x\le 2\pi
Isolating cos(x)
cos(x) = -1/2
General solutions for cos(x) = -1/2#
\cos(x)=-\frac{1}{2}:\quad x=\frac{2\pi }{3}+2\pi n,quad x=\frac{4\pi }{3}+2\pi n
Solutions for the range 0\le x\le 2\pi
x=\frac{2\pi }{3},x=\frac{4\pi }{3}
Finally combining all the solutions,
x=\frac{2\pi }{3},x=\frac{\pi }{4},x=\frac{3\pi }{4},x=\frac{5\pi }{4},x=\frac{4\pi }{3},x=\frac{7\pi }{4}
