Solve:

2cos²x + 3sinx = 0

2(1 − sin²x) + 3sin(x) = 0

⇒ 2sin²x − 3sin(x) − 2 = 0

⇒ (2sinx + 1)(sinx − 2) = 0

⇒ sinx = −1/2 or sin(x) = 2

⇒ sinx = 2 is not possible (Why?)

⇒ sinx = −1/2 = sin(7π/6)

Hence, the solution is given by

x = nπ + (−1)ⁿ·(7π/6), n ∈ ℤ

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