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How do I find the exact value of sin2x when secx is equal to the negative square root of three

  • trig -

    then cosx = -1/√3

    x must be in II or III

    in II:
    sinx = √2/√3
    sin 2x = 2sinxcosx = 2(√2/√3)(-1/√3) = -2√2/3

    in III:
    sinx = -√2/√3
    sin 2x = +2√2/3

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