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Math Help - [SOLVED] A challenging trigonometric equation?

  1. #1
    Super Member fardeen_gen's Avatar
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    [SOLVED] A challenging trigonometric equation?

    Solve √(√3 cos x + sin x - 2) + √(cot 3x + sin^2 x - 1/4) = sin (3x/2) + √2/2?

    ^- raised to
    Answer: (24n + 13)π/6 , n belongs to Z

    How to solve this? My answer is not matching.

    Any help would be appreciated.
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  2. #2
    MHF Contributor red_dog's Avatar
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    The equation can be written as
    \displaystyle\sqrt{2}\sqrt{\frac{\sqrt{3}}{2}\cos x+\frac{1}{2}\sin x-1}+\sqrt{\cot 3x+\sin^2x-\frac{1}{4}}=\sin\frac{3x}{2}+\frac{\sqrt{2}}{2}\L  eftrightarrow
    \displaystyle\Leftrightarrow\sqrt{2}\sqrt{\sin\lef  t(x+\frac{\pi}{3}\right)-1}+\sqrt{\cot 3x+\sin^2x-\frac{1}{4}}=\sin\frac{3x}{2}+\frac{\sqrt{2}}{2}
    We have to put the condition \sin\left(x+\frac{\pi}{3}\right)\geq 1. But \sin\left(x+\frac{\pi}{3}\right)\leq 1, \ \forall x\in\mathbf{R}.
    Then \displaystyle\sin\left(x+\frac{\pi}{3}\right)=1\Ri  ghtarrow x=\frac{\pi}{6}+2k\pi, \ k\in\mathbf{Z}.
    Now \displaystyle \cot3\left(\frac{\pi}{6}+2k\pi\right)=0, \ \sin^2\left(\frac{\pi}{6}+2k\pi\right)=\frac{1}{4}
    so, the left hand member of the equation is equal to 0.
    Then \displaystyle \sin\frac{3}{2}\left(\frac{\pi}{6}+2k\pi\right)=-\frac{\sqrt{2}}{2}\Rightarrow x=\frac{(24p+13)\pi}{6}, \ p\in\mathbf{Z}
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  3. #3
    Super Member fardeen_gen's Avatar
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    Can anyone explain the last step?
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  4. #4
    MHF Contributor red_dog's Avatar
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    \displaystyle\sin\left(\frac{\pi}{4}+3k\pi\right)=-\frac{\sqrt{2}}{2}\Rightarrow\sin\left(\frac{\pi}{  4}+k\pi\right)=-\frac{\sqrt{2}}{2}
    Then k must be odd: k=2p+1
    Now \displaystyle x=\frac{\pi}{6}+2k\pi=\frac{\pi}{6}+2(2p+1)\pi=\fr  ac{(24p+13)\pi}{6}
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