Test:333 QNo:89

The number of roots of the equation 1+cos2θ=sin3θ in the interval [0,2π] is

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11
virang1 Jhaveri ·

2-sin2θ = sin3θ
sin3θ + sin2θ -2 = 0

(Sinθ -1)*(Sin2θ + 2Sinθ +2)
Therefore 1 real and 2 imaginary roots

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