Решите неравенство sin2x-cosx<0

$sin2x - cosx \ \textless \ 0, \\ 2sinxcosx - cosx \ \textless \ 0, \\ cosx(2sinx-1)\ \textless \ 0.\\ \\ f(x)=cosx(2sinx-1);\\f(x) = 0, \\ cosx(2sinx-1) = 0. \\ \\ cosx = 0, \\ x = \frac{ \pi }{2} + \pi n. \\ \\sinx = \frac{1}{2}, \\ x = \frac{ \pi }{6} + 2 \pi n, x= \frac{5 \pi }{6} + 2 \pi n. \\ \\ sin2x - cosx \ \textless \ 0, \\ -\frac{ \pi }{2} + 2 \pi n \ \textless \ x \ \textless \ \frac{ \pi }{6} + 2 \pi n, \\ \frac{ \pi }{2} + 2 \pi n\ \textless \ x \ \textless \ \frac{5 \pi }{6} + 2 \pi n .$

Примечание: n ∈ Z.