Помогите решить 16ое неравенство!

$cos^{2} 2x\ \textgreater \ \frac{2+ \sqrt{3} }{4} \\ \frac{1+cos4x}{2} \ \textgreater \ \frac{2+ \sqrt{3} }{4}|*4 \\ 2+2cos4x\ \textgreater \ 2+ \sqrt{3} \\ cos4x\ \textgreater \ \frac{ \sqrt{3} }{2} \\ cos4x= \frac{ \sqrt{3} }{2} \\ 4x= \left \{ {{+} \atop {-}} \right. \frac{ \pi }{6} +2 \pi n; \\ x= \left \{ {{+} \atop {-}} \right. \frac{ \pi }{24} + \frac{ \pi n}{2}$
,n∈Z
$-\frac{ \pi }{24} + \frac{ \pi n}{2} \ \textless \ x\ \textless \ \frac{ \pi }{24} + \frac{ \pi n}{2}$
отв:x∈(-π/24 + πn/2;π/24 + πn/2),n∈Z