[tex]2\sin^2x-\cos2x=2\sin^2x-(2\cos^2x-1)=2(\sin^2-\cos^2x)+1=0[/tex]
[tex]-2\cos2x=-1[/tex]
[tex]\cos2x=\dfrac12[/tex]
You have [tex]\cos x=\dfrac12[/tex] for
[tex]x=\dfrac\pi3+2n\pi[/tex]
[tex]x=-\dfrac\pi3+2n\pi[/tex]
which means [tex]\cos 2x=\dfrac12[/tex] for
[tex]2x=\dfrac\pi3+2n\pi\implies x=\dfrac\pi6+n\pi[/tex]
[tex]2x=-\dfrac\pi3+2n\pi\implies x=-\dfrac\pi6+n\pi[/tex]
where [tex]n[/tex] is any integer.
