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