решите что нибудь хотя бы С1

$6sin^2 x+7cos x-1=0,$

$6(1-cos^2 x)+7cos x-1=0,$

$-6cos^2 x+7cos x+5=0,$

$6cos^2 x-7cos x-5=0,$

cos x=t,

6t²-7t-5=0,

D=169,

t1=-½, t2=1 ⅔>1,

cos x=-½,

x=±arccos (-½) + 2πk, k∈Z,

x=±(π-arccos (½)) + 2πk, k∈Z,

x=±(π-π/3) + 2πk, k∈Z,

x=±2π/3 + 2πk, k∈Z,

-7π/2<±2π/3 + 2πk<-5π/2,</p>

[-7π/2-2π/3<2πk<-5π/2-2π/3, -7π/2+2π/3<2πk<-5π/2+2π/3,</p>

[-25π/6<2πk<-19π/6, -17π/6<2πk<-11π/6,</p>

[-25/12

[k=-2, k=-1,

x=2π/3 - 4π,

x=-10π/3;

x=-2π/3-2π,

x=-8π/3.