# Solving a trig equation - Is my work correct?

• Nov 28th 2009, 07:52 AM
iluvmathbutitshard
Solving a trig equation - Is my work correct?
Find all solutions of the equation in the interval [0, 2pi):
sin (x + pi/6) - sin (x - pi/6) = 1/2

My work:
sin x cos pi/6 + cos x sin pi/6 - sin x cos pi/6 - cos x sin pi/6 = 1/2
2 sin x cos pi/6 = 1/2
(2 sin x) (sqrt3/2) = 1/2
Now this is the part where I am confused,
2 sin x = (1/2) / (sqrt3/2)
2 sin x = (2) / (2 sqrt3)
2 sin x = 1/sqrt 3
sin x = sqrt3/2
x = pi/3, 2pi/3

Is this right?
Any help is appreciated
• Nov 28th 2009, 09:26 AM
masters
Quote:

Originally Posted by iluvmathbutitshard
Find all solutions of the equation in the interval [0, 2pi):
sin (x + pi/6) - sin (x - pi/6) = 1/2

My work:
sin x cos pi/6 + cos x sin pi/6 - sin x cos pi/6 - cos x sin pi/6 = 1/2
2 sin x cos pi/6 = 1/2
(2 sin x) (sqrt3/2) = 1/2
Now this is the part where I am confused,
2 sin x = (1/2) / (sqrt3/2)
2 sin x = (2) / (2 sqrt3)
2 sin x = 1/sqrt 3
sin x = sqrt3/2
x = pi/3, 2pi/3

Is this right?
Any help is appreciated

Hi iluvmathbutitshard,

I think you have a sign wrong in your first step:

$\displaystyle \sin x \cos \frac{\pi}{6}+\cos x \sin \frac{\pi}{6}-(\sin x \cos \frac{\pi}{6}-\cos x \sin \frac{\pi}{6})=\frac{1}{2}$

$\displaystyle \sin x \cos \frac{\pi}{6}+\cos x \sin \frac{\pi}{6}-\sin x \cos \frac{\pi}{6}+\cos x \sin \frac{\pi}{6}=\frac{1}{2}$

$\displaystyle \frac{\sqrt{3}}{2}\sin x+\frac{1}{2} \cos x -\frac{\sqrt{3}}{2} \sin x+\frac{1}{2} \cos x=\frac{1}{2}$

$\displaystyle \cos x = \frac{1}{2}$

$\displaystyle x=\left\{\frac{\pi}{3}, \frac{5\pi}{3}\right\}$