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Thread: solve this trig equation

  1. April 6th 2010, 03:38 PM #1
    magmagod
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    solve this trig equation

    sin2x + sqrt 2/2 = 0

    i moved sqrt 2/2 to the other side..then i found the inverse of sin for 2x...then i got the quadrant values, dividing each one by 2...after than i added the period (pi/2) to the results giving the final values for x or theta..so is what i did right?
    Last edited by magmagod; April 6th 2010 at 03:59 PM.
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  2. April 6th 2010, 04:42 PM #2
    harish21
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    Quote Originally Posted by magmagod View Post
    sin2x + sqrt 2/2 = 0

    i moved sqrt 2/2 to the other side..then i found the inverse of sin for 2x...then i got the quadrant values, dividing each one by 2...after than i added the period (pi/2) to the results giving the final values for x or theta..so is what i did right?
    sin2x + \frac{\sqrt{2}}{2} = 0

    sin2x = - \frac{\sqrt{2}}{2}

    2x = sin^{-1}( \frac{\sqrt{2}}{2})

    2x = \frac{-\pi}{4}

    x = \frac{-\pi}{8}
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