Question

Решите неравенство sin2x-cosx<0

Answer 1

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

Примечание: n ∈ Z.