2-sin2θ = sin3θ
sin3θ + sin2θ -2 = 0
(Sinθ -1)*(Sin2θ + 2Sinθ +2)
Therefore 1 real and 2 imaginary roots
The number of roots of the equation 1+cos2θ=sin3θ in the interval [0,2π] is
2-sin2θ = sin3θ
sin3θ + sin2θ -2 = 0
(Sinθ -1)*(Sin2θ + 2Sinθ +2)
Therefore 1 real and 2 imaginary roots