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=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