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}#
