2sin2x+cosx+4sinx+1=0 cos4x+4sin²x=1+2sin²2x 2cosx/2sin3x=cosx/2 1/tg²2x - 2/tgx - 3 = 0 решить что-нибудь : )

2sin2x+cosx+4sinx+1=0

4sin(x)+cos(x)+4sin(x)*cos(x)+1=0

(4sin(x)+1)(cos(x)+1)=0

x=2πn-π, n∈Z

x=2πn+π, n∈Z

x=2πn-arcsin(1/4), n∈Z

x=2πn+π+arcsin(1/4), n∈Z

cos4x+4sin²x=1+2sin²2x

4sin²(x)+cos(4x)=2-cos(4x)

cos(2x)=cos(4x)

x=πn, n∈Z

x=-π/3+πn, n∈Z

x=π/3+πn, n∈Z

cos4x=1-2sin^2(2x) => 1-2sin^2(2x)+4sin^2(x)=1+2sin^2(2x) =>

4sin^2(x)=4sin^2(2x) => sin^2(2x)=4sin^2(x)*cos^2(x) =>

cos^2(x)=1/4 => cos(x)=1/2 =

x=arccos(1/2)+2πk    k€Z

x= π /3+2 πk   k€Z