text: # Thread: Solve this trig equation
  
  1. ## Solve this trig equation
  
  sin(3x) = cos(x + pi/4)
  
  Thanks
  
  2. 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
  
  3. $\displaystyle \sin(3x) = \cos( x + \frac{\pi}{4} )$
  
  $\displaystyle \sin(3x) = \sin( \frac{\pi}{2} - x - \frac{\pi}{4} )$
  
  $\displaystyle \sin(3x) = \sin( \frac{\pi}{4} - x )$
  
  $\displaystyle 3x = 2n\pi + \frac{\pi}{4} - x$
  
  $\displaystyle x = \frac{(8n+1)\pi}{16}$
  
  Or
  
  $\displaystyle 3x = (2n+1)\pi - \frac{\pi}{4} + x$
  
  $\displaystyle x = \frac{( 8n + 3 )\pi}{8}$
  
  4. where did 2 n come from?
  
  5. Originally Posted by usagi_killer
  where did 2 n come from?
  its because there is no restriction to the domain of x .
  
  6. 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