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13 June, 15:31

Find the Solution of 2cos^2x + 3sin x = 0

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  1. 13 June, 18:52
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    2cos^2 x + 3sin x = 0

    2 (1 - sin^2 x) + 3sin x = 0

    2 - 2sin^2 x + 3sin x = 0

    2sin^2 x - 3sin x - 2 = 0

    Let sin x = m, then

    2m^2 - 3m - 2 = 0

    2m^2 + m - 4m - 2 = 0

    m (2m + 1) - 2 (2m + 1) = 0

    (m - 2) (2m + 1) = 0

    m = 2 or m = - 1/2

    Now, sin x = - 1/2

    Therefore, x = 1/6 (12nπ - π) and 1/6 (12nπ + 7π)
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