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Math Help - Solve this trig equation

  1. #1
    Senior Member
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    Solve this trig equation

    sin(3x) = cos(x + pi/4)

    Thanks
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  2. #2
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    my attempt
    sin 3x = cos(x+pi/4)
    sin 3x= sin{ pi/2- (x+pi/2)}
    sin 3x= sin( pi/4 - x )
    gives
    3x=pi/4 - x

    x=pi/16
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  3. #3
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     \sin(3x)  = \cos( x + \frac{\pi}{4} )

     \sin(3x) = \sin( \frac{\pi}{2}  -  x - \frac{\pi}{4} )

     \sin(3x) = \sin( \frac{\pi}{4} - x )


     3x = 2n\pi + \frac{\pi}{4} - x

     x = \frac{(8n+1)\pi}{16}

    Or

     3x = (2n+1)\pi - \frac{\pi}{4} + x

     x = \frac{( 8n + 3 )\pi}{8}
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  4. #4
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    where did 2 n come from?
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  5. #5
    MHF Contributor
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    Quote Originally Posted by usagi_killer View Post
    where did 2 n come from?
    its because there is no restriction to the domain of x .
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  6. #6
    Senior Member pacman's Avatar
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    i have plotted this function, sin(3x) - cos(x + pi/4) = 0, for x = 0 to 2pi

    it does have 6 roots for x = 0 to 2 pi

    using the sum to product formula, we have this

    -2 sin (pi/8 -2x) cos (x + pi/8) = 0

    this one form does not generate the 6 roots as shown in the plot, i wonder
    Attached Thumbnails Attached Thumbnails Solve this trig equation-wert.gif  
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