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Math Help - Fantastic Trigo-equation

  1. #1
    Super Member dhiab's Avatar
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    Fantastic Trigo-equation

    Solve :
    \left\{ \begin{array}{l}<br />
\sin x = \frac{{\sqrt {7 - \sqrt {21 + \sqrt {80} } } }}{{1 + \sqrt {7 + \sqrt {48} } - \sqrt {4 - \sqrt {12} } }} \\ <br />
0 < x < \frac{\pi }{2} \\ <br />
\end{array} \right.
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  2. #2
    MHF Contributor red_dog's Avatar
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    \sqrt{21+\sqrt{80}}=\sqrt{(\sqrt{20}+1)^2}=\sqrt{2  0}+1

    \sqrt{7-\sqrt{21+\sqrt{80}}}=\sqrt{6+\sqrt{20}}=\sqrt{(1+\  sqrt{5})^2}=1+\sqrt{5}

    \sqrt{7+\sqrt{48}}=\sqrt{(2+\sqrt{3})^2}=2+\sqrt{3  }

    \sqrt{4-\sqrt{12}}=\sqrt{(\sqrt{3}-1)^2}=\sqrt{3}-1

    Then the equation is

    \sin x=\frac{1+\sqrt{5}}{4}

    We have \sin\frac{3\pi}{10}=\cos\left(\frac{\pi}{2}-\frac{3\pi}{10}\right)=\cos\frac{2\pi}{10}=1-2\sin^2\frac{\pi}{10}

    3\sin\frac{\pi}{10}-4\sin^3\frac{\pi}{10}=1-2\sin^2\frac{\pi}{10}

    Let \sin\frac{\pi}{10}=t\Rightarrow 4t^3-3t^2-3t+1=0\Rightarrow t_1=1, \ t_2=\frac{\sqrt{5}-1}{4}, \ t_3=\frac{-1-\sqrt{5}}{4}

    But \sin\frac{\pi}{10}\in(0,1)\Rightarrow \sin\frac{\pi}{10}=\frac{\sqrt{5}-1}{4}

    Then \sin\frac{3\pi}{10}=1-2\sin^2\frac{\pi}{10}=1-\left(\frac{\sqrt{5}-1}{4}\right)^2=\frac{1+\sqrt{5}}{4}

    Therefore x=\frac{3\pi}{10}
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  3. #3
    Super Member dhiab's Avatar
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    Quote Originally Posted by red_dog View Post
    \sqrt{21+\sqrt{80}}=\sqrt{(\sqrt{20}+1)^2}=\sqrt{2  0}+1

    \sqrt{7-\sqrt{21+\sqrt{80}}}=\sqrt{6+\sqrt{20}}=\sqrt{(1+\  sqrt{5})^2}=1+\sqrt{5}

    \sqrt{7+\sqrt{48}}=\sqrt{(2+\sqrt{3})^2}=2+\sqrt{3  }

    \sqrt{4-\sqrt{12}}=\sqrt{(\sqrt{3}-1)^2}=\sqrt{3}-1

    Then the equation is

    \sin x=\frac{1+\sqrt{5}}{4}

    We have \sin\frac{3\pi}{10}=\cos\left(\frac{\pi}{2}-\frac{3\pi}{10}\right)=\cos\frac{2\pi}{10}=1-2\sin^2\frac{\pi}{10}

    3\sin\frac{\pi}{10}-4\sin^3\frac{\pi}{10}=1-2\sin^2\frac{\pi}{10}

    Let \sin\frac{\pi}{10}=t\Rightarrow 4t^3-3t^2-3t+1=0\Rightarrow t_1=1, \ t_2=\frac{\sqrt{5}-1}{4}, \ t_3=\frac{-1-\sqrt{5}}{4}

    But \sin\frac{\pi}{10}\in(0,1)\Rightarrow \sin\frac{\pi}{10}=\frac{\sqrt{5}-1}{4}

    Then \sin\frac{3\pi}{10}=1-2\sin^2\frac{\pi}{10}=1-\left(\frac{\sqrt{5}-1}{4}\right)^2=\frac{1+\sqrt{5}}{4}

    Therefore x=\frac{3\pi}{10}
    hello Thank you
    I'thinque in the second line you are a error calculs .
    LOOK HERE :
    Attached Thumbnails Attached Thumbnails Fantastic Trigo-equation-17.jpg  
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