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Thread: Solve the following equations

  1. #1
    dno
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    Solve the following equations

    cos2(theta) = 1 - cos(theta) [-180, 180]

    sin^2(theta) = 2sin2(theta) [-180,180]

    2tan(theta) = sqrt3(1 - tan(theta)(1 + tan(theta)) [0,360]


    thanks, i keep getting the wrong answers
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  2. #2
    MHF Contributor
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    Quote Originally Posted by dno View Post
    cos2(theta) = 1 - cos(theta) [-180, 180]

    sin^2(theta) = 2sin2(theta) [-180,180]

    2tan(theta) = sqrt3(1 - tan(theta)(1 + tan(theta)) [0,360]
    $\displaystyle \cos^2{t} = 1 - \cos{t}$

    $\displaystyle \cos^2{t} + \cos{t} - 1 = 0$

    $\displaystyle \cos{t} = \frac{-1 \pm \sqrt{5}}{2}$


    $\displaystyle \sin^2{t} = 2\sin(2t)$

    $\displaystyle \sin^2{t} = 4\sin{t}\cos{t}$

    $\displaystyle \sin^2{t} - 4\sin{t}\cos{t} = 0$

    $\displaystyle \sin{t}(\sin{t} - 4\cos{t}) = 0$

    $\displaystyle \sin{t} = 0$

    $\displaystyle \sin{t} = 4\cos{t}$

    $\displaystyle \tan{t} = 4$


    $\displaystyle 2\tan{t} = \sqrt{3}(1 - \tan{t})(1 + \tan{t})$

    $\displaystyle 2\tan{t} = \sqrt{3}(1 - \tan^2{t})$

    $\displaystyle 2\tan{t} = \sqrt{3} - \sqrt{3}\tan^2{t}$

    $\displaystyle \sqrt{3}\tan^2{t} + 2\tan{t} - \sqrt{3} = 0$

    $\displaystyle (\sqrt{3}\tan{t} - 1)(\tan{t} + \sqrt{3}) = 0$

    $\displaystyle \tan{t} = \frac{1}{\sqrt{3}}$

    $\displaystyle \tan{t} = -\sqrt{3}$
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