Question

1)6cos^2x-cosx-2/корень из -sin=02)корень из cos2x=1+2sinx3)sinx+sin2x/sin3x=-1помогите решить

Answer 1

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