1)6cos^2x-cosx-2/корень из -sin=0
2)корень из cos2x=1+2sinx
3)sinx+sin2x/sin3x=-1
помогите решить

1)sinx<0⇒x∈(π+2πn;2π+2πn)6cos²x-cosx-2=0cosx=a6a²-a-2=0D=1+48=49a1=(1-7)/12=-1/2⇒cosx=-1/2⇒x=-2π/3+2πn U x=2π/3+2πnx=4π/3+2πn∈(π+2πn;2π+2πn)a2=(1+7)/12=2/3⇒cosx=-arccos2/3+2πn U x=arccoax+2πnx=3/2+-accos2/3+2πn∈(π+2πn;2π+2πn)2)1+2sinx≥0⇒sinx≥-1/2⇒x∈[-π/3+2πn;4π/3+2πn]cos2x=1+4sinx+4sin²x1-2sin²x=1+4sinx+4sin²x6sin²x+4sinx=02sinx(3sinx+2)=0sinx=0⇒x=πnsinx=-2/3 x=(-1)^n+1*arcsinx+πn∉[-π/3+2πn;4π/3+2πn]3)sin3x≠0⇒x≠πn/32sin(3x/2)cos(x/2)/2sin(3x/2)cos(3x/2)=-1cos(x/2)/cos(3x/2)=-1cos(x/2)+cos(3x/2)=02cosxcos(x/2)=0cosx=0⇒x=π/2+πncosx/2=0⇒x/2=π/2+πn⇒x=π+2πn