[tex]\cos^2x-\sin^2x=\cos2x=\dfrac{\sqrt3}2[/tex]
In the interval [tex][0,2\pi)[/tex], [tex]\cos x[/tex] takes on the value of [tex]\dfrac{\sqrt3}2[/tex] when [tex]x=\pm\dfrac\pi3[/tex]. In general, this happens for [tex]x=\pm\dfrac\pi3+2n\pi[/tex] where [tex]n[/tex] is any integer.
So, the solutions to this equation satisfy
[tex]2x=\pm\dfrac\pi3+2n\pi[/tex]
or, dividing by 2,
[tex]x=\pm\dfrac\pi6+n\pi[/tex]
for all integers [tex]n[/tex].
