# solve this trig equation

• Apr 6th 2010, 04:38 PM
magmagod
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?
• Apr 6th 2010, 05:42 PM
harish21
Quote:

Originally Posted by magmagod
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?

$sin2x + \frac{\sqrt{2}}{2} = 0$

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

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

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

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