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29 July, 17:18

Solve sin2x + 4sin x + 3 = 0

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  1. 29 July, 17:44
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    Sin² x + 4 sin x + 3 = 0

    Substitution: u = sin x

    u² + 4 u + 3 = 0

    u² + 3 u + u + 3 = 0

    u (u + 3) + (u + 3) = 0

    (u + 3) (u + 1) = 0

    u = - 3, u = - 1;

    sin x = - 3 (false)

    sin x = - 1

    Answer:

    x = 3 π / 2 + 2 k π, k ∈ Z
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