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Thread: Trig Identities

  1. #1
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    Trig Identities

    Solve for $\displaystyle 0\leq x \leq 2\pi$

    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 $\displaystyle 0\leq x \leq 2\pi$

    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 $\displaystyle \cos(2x) = 1 - 2\sin^2x$.

    Get:

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

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

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

    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 $\displaystyle \pi/6$
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  4. #4
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    Yes $\displaystyle \pi/6$ is a solution.
    There are several answers in that interval given $\displaystyle (0 \le x \le 2\pi)$.

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

    Do you see why?
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