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Math Help - triq eqautions

  1. #1
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    triq eqautions

    cos2t+2cos^2(t/2)-2=0
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  2. #2
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    Quote Originally Posted by Thelema View Post
    cos2t+2cos^2(t/2)-2=0
    \cos 2t+2\cos^2(\frac{t}2)-2=0

    \cos 2t+(2\cos^2(\frac{t}2)-1) -1=0

    \cos 2t+\cos t -1=0

    2\cos^2 t - 1 +\cos t -1=0

    2\cos^2 t +\cos t - 2=0

    I am sure you know what to do now
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  3. #3
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    Quote Originally Posted by Thelema View Post
    cos2t+2cos^2(t/2)-2=0
    Double angle formula: \cos (2A) = 2 \cos^2 (A) - 1.

    Substitute A = t/2: \cos (t) = 2 \cos^2 (t/2) - 1 \Rightarrow \cos^2 (t/2) = \frac{1}{2} \, (1 + \cos t).

    So your equation can be re-written as

    (2 \cos^2 t - 1) + (1 + \cos t) - 2 = 0 \Rightarrow 2 \cos^2 t + \cos t - 2 = 0 ......

    Check for mistakes - but you get the idea, right?
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  4. #4
    Super Member wingless's Avatar
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    \cos 2t + 2\cos^2 \frac{t}{2} - 2 = 0

    \cos 2t + 2\cos^2 \frac{t}{2} = 2

    \cos^2 t - \sin^2 t + 2\cos^2 \left(\frac{t}{2}\right) = 2

    ..... \cos^2 \frac{t}{2} = \frac{1+\cos t}{2}.....

    \cos^2 t - \sin^2 t + \not{2}.\frac{1+\cos t}{\not 2} = 2

    \cos^2 t - \sin^2 t + 1 + \cos t = 2

    ..... 1 = \cos^2 t + \sin^2 t.....

    2\cos^2 t + \cos t = 2

    Let a= \cos t..

    2a^2+ a -2 = 0

    a = \cos t = \frac{\sqrt{17}-1}{4},\frac{-\sqrt{17}-1}{4}

    \cos t = \frac{-\sqrt{17}-1}{4} is not possible because cos is between -1 and +1.

    So we only use
    \cos t = \frac{\sqrt{17}-1}{4}

    Solve this for t.
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