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24 February, 06:20

Given that 2sin^2 theta + sin theta - 1 = 0, find the two values for sin theta.

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  1. 24 February, 06:45
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    2sin^2 θ + sin θ - 1 = 0

    2sin^2 θ + 2sin θ - sin θ - 1 = 0

    2sin θ (sin θ + 1) - 1 (sin θ + 1) = 0

    (2sin θ - 1) (sin θ + 1) = 0

    2sin θ - 1 = 0 or sin θ + 1 = 0

    sin θ = 1/2 or sin θ = - 1

    The two values of sin θ are 1/2 and - 1
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