2cos^2x-3sinxcosx+sin^2x=0 и sin(пи/6+2x)+1=0 решите !

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1)
2cos²x - 3sinxcosx + sin²x=0

Делим на cos²x:

2 - 3tgx + tg²x=0
tg²x -3tgx +2=0

y=tgx
y² -3y +2=0
D=9-8=1
y₁=(3-1)/2=1
y₂=(3+1)/2=2

При y=1
tgx=1
x=π/4 + πk,  k∈Z

При y=2
tgx=2
x=arctg2+πn, n∈Z

ответ: π/4+πk, k∈Z;
            arctg2 + πn, n∈Z.

2)
sin(π/6 +2x)= -1
π/6+2x= -(π/2) + 2πk
2x= -(π/6) - (π/2) +2πk
2x= -(2π/3) +2πk
x= -π/3 + πk, k∈Z
ответ: -π/3 + πk,  k∈Z.