Доказать тождество:2sin^2(L)+cos(L)=1

$2sin ^{2} x+cosx=1\\2sin ^{2} x+cosx=sin^{2} x+cos ^{2} x\\2sin ^{2} x+cosx-sin^{2} x-cos ^{2} x =0 \\ sin^{2} x-cos^{2} x+cosx=0 \\ 1-cos^{2} x-cos^{2} x+cosx=0 \\ 1-2cos^{2} x+cosx=0 /*(-1)\\ 2cos^{2} x-cosx-1=0 \\ cosx=t \\ 2t ^{2} -t-1=0 \\ D=1+8=9 \\$
$\sqrt{D} =3 \\ t _{1} = \frac{1+3}{4} =1 \\ t _{2} = \frac{1-3}{4} =- \frac{1}{2} \\ cosx=1 \\x=2 \pi n \\ cosx=- \frac{1}{2} \\ x=+- \frac{2 \pi }{3} +2 \pi n$

n∈Z