3 sin^2x-cosx+1=0 Помогите решить

[latex]3 sin^2x-cosx+1=0 [/latex] [latex]3 sin^2x+1-cosx=0 [/latex] [latex]3 sin^2x+2sin^2 \frac{x}{2} =0[/latex] [latex]3*4 sin^2 \frac{x}{2}cos^2 \frac{x}{2} +2sin^2 \frac{x}{2} =0[/latex] [latex]12 sin^2 \frac{x}{2}cos^2 \frac{x}{2} +2sin^2 \frac{x}{2} =0[/latex] [latex]6 sin^2 \frac{x}{2}cos^2 \frac{x}{2} +sin^2 \frac{x}{2} =0[/latex] [latex] sin^2 \frac{x}{2}(6cos^2 \frac{x}{2} +1) =0[/latex] [latex] sin^2 \frac{x}{2}=0[/latex]          или       [latex]6cos^2 \frac{x}{2} +1 =0[/latex] [latex] sin \frac{x}{2} =0[/latex]            или       [latex]cos^2 \frac{x}{2} =- \frac{1}{6} [/latex] [latex] \frac{x}{2} = \pi n,[/latex] [latex]n[/latex] ∈ [latex]Z[/latex]    или       ∅ [latex]x} = 2\pi n,[/latex] [latex]n[/latex] ∈ [latex]Z[/latex] [latex]sin^2 \frac{x}{2} = \frac{1-cosx}{2} [/latex] ⇒   [latex]2sin^2 \frac{x}{2}=1-cosx[/latex] [latex]sin2x=2sinxcosx[/latex] ⇒   [latex]sinx=2sin \frac{x}{2} cos \frac{x}{2} [/latex]