3 sin^2x-cosx+1=0 Помогите решить

$3 sin^2x-cosx+1=0$
$3 sin^2x+1-cosx=0$
$3 sin^2x+2sin^2 \frac{x}{2} =0$
$3*4 sin^2 \frac{x}{2}cos^2 \frac{x}{2} +2sin^2 \frac{x}{2} =0$
$12 sin^2 \frac{x}{2}cos^2 \frac{x}{2} +2sin^2 \frac{x}{2} =0$
$6 sin^2 \frac{x}{2}cos^2 \frac{x}{2} +sin^2 \frac{x}{2} =0$
$sin^2 \frac{x}{2}(6cos^2 \frac{x}{2} +1) =0$
$sin^2 \frac{x}{2}=0$ или    $6cos^2 \frac{x}{2} +1 =0$
$sin \frac{x}{2} =0$     или    $cos^2 \frac{x}{2} =- \frac{1}{6}$
$\frac{x}{2} = \pi n,$ $n$ ∈ $Z$ или   ∅
$x} = 2\pi n,$ $n$ ∈ $Z$

$sin^2 \frac{x}{2} = \frac{1-cosx}{2}$ ⇒ $2sin^2 \frac{x}{2}=1-cosx$
$sin2x=2sinxcosx$ ⇒ $sinx=2sin \frac{x}{2} cos \frac{x}{2}$