# 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