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Thread: Trig equations

  1. September 4th 2009, 11:03 PM #1
    requal
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    Trig equations

    Find the general solution of equation sin2x=cosx
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  2. September 4th 2009, 11:07 PM #2
    red_dog
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    \sin 2x=\sin\left(\frac{\pi}{2}-x\right)

    \sin 2x-\sin\left(\frac{\pi}{2}-x\right)=0

    Now use the identity \sin a-\sin b=2\sin\frac{a-b}{2}\cos\frac{a+b}{2}
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  3. September 5th 2009, 12:01 AM #3
    skeeter
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    Quote Originally Posted by requal View Post
    Find the general solution of equation sin2x=cosx
    \sin(2x) = \cos{x}

    2\sin{x}\cos{x} = \cos{x}

    2\sin{x}\cos{x} - \cos{x} = 0

    \cos{x}(2\sin{x} - 1) = 0

    set each factor equal to zero and solve.
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  4. September 5th 2009, 02:08 PM #4
    Amer
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    Quote Originally Posted by requal View Post
    Find the general solution of equation sin2x=cosx

    \sin 2x=\cos x

    2\sin x\cos x = \cos x

    2\sin x \cos x - \cos x =0

    \cos x(2\sin x -1 )=0

    \cos x = 0 \Rightarrow x=\frac{\pi}{2} + n\pi

    \sin x =\frac{1}{2} \Rightarrow x=\frac{\pi}{6} + 2n\pi and x=\frac{5\pi}{6}+2n\pi
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