Question

& Find the general solution of equation sin2n+cosn=0​

Answer 1

Answer:

Therefore, the general solution of the equation sin 2θ = 0 is θ = nπ2, where, n = 0, ± 1, ± 2, ± 3,……. ⇒ x = 2nπ3, where, n = 0, ± 1, ± 2, ± 3,……. Therefore, the general solution of the equation sin 3x2 = 0 is θ = 2nπ3, where, n = 0, ± 1, ± 2, ± 3,……

Step-by-step explanation:

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