# Solve this trig equation

• Sep 16th 2009, 01:14 AM
usagi_killer
Solve this trig equation
sin(3x) = cos(x + pi/4)

Thanks
• Sep 16th 2009, 01:23 AM
ramiee2010
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
• Sep 16th 2009, 01:44 AM
simplependulum
$\displaystyle \sin(3x) = \cos( x + \frac{\pi}{4} )$

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

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

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

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

Or

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

$\displaystyle x = \frac{( 8n + 3 )\pi}{8}$
• Sep 16th 2009, 02:02 AM
usagi_killer
where did 2 n come from?
• Sep 16th 2009, 03:28 AM
Quote:

Originally Posted by usagi_killer
where did 2 n come from?

its because there is no restriction to the domain of x .
• Sep 18th 2009, 02:35 AM
pacman
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