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Math Help - Solving for theta

  1. #1
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    Solving for theta

    Solve sin^2ϴ = 1 + cosϴ 0 ≤ ϴ ≤ 2pi


    I dont even know where to start! any help would be appreciated thank you
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  2. #2
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    e^(i*pi)'s Avatar
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    Quote Originally Posted by cordelljheckel View Post
    Solve sin^2ϴ = 1 + cosϴ 0 ≤ ϴ ≤ 2pi


    I dont even know where to start! any help would be appreciated thank you
    sin^2 \theta = 1-cos^2 \theta

    Sub in, 1 will cancel and solve the quadratic
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  3. #3
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    so am i substituting sin^2(x) with 1 - cos2x / 2 then solving?
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  4. #4
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    No, that wasn't the given indication. Read it again.
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  5. #5
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    i dont understand, which identity are you using to sub in?
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  6. #6
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    Quote Originally Posted by cordelljheckel View Post
    i dont understand, which identity are you using to sub in?
    Just like e^(i*pi) said,

    sin^2(\theta) = 1 - cos^2(\theta)

    Substitute into the first equation to get:

    1-cos^2\theta = 1 + cos\theta

    Now solve.
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