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Math Help - Trig Identities

  1. #1
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    Trig Identities

    Solve for 0\leq x \leq 2\pi

    well i am bad at the math tool, but its 0<=x<=2pi

    Cos2x = 1 - sinx

    What would be the best way to set this up?
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  2. #2
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    Quote Originally Posted by adam182 View Post
    Solve for 0\leq x \leq 2\pi

    well i am bad at the math tool, but its 0<=x<=2pi

    Cos2x = 1 - sinx

    What would be the best way to set this up?
    Hi Adam.

    Use the identity that \cos(2x) = 1 - 2\sin^2x.

    Get:

    1-2\sin^2x = 1 - \sin x.

    So 2 \sin^2 x - \sin x = 0.

    Factor:
    \sin x \cdot (2 \sin x - 1) = 0.

    Can you finish it from here?
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  3. #3
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    Yes, I got to x=30 which I converted to radians...so \pi/6
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  4. #4
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    Yes \pi/6 is a solution.
    There are several answers in that interval given (0 \le x \le 2\pi).

    Note that 5\pi/6 is also an answer, as are 0, \pi, and 2\pi.

    Do you see why?
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