Question

4sin2x-cosx-1=0 решения

Answer 1

$4sin^2x-cosx-1=0\4-4cos^2x-cosx-1=0\4cos^2x+cosx-3=0\cosx_{1,2}=frac{-1^+_-7}{8}\cosx=frac{3}{4} cosx=-1\x_{1,2}=^+_-arccosfrac{3}{4}+2pi n; x_3=pi+2pi n;nin Z;$