# Help with trigonometric equation

• Nov 8th 2009, 09:10 AM
LnVII
Help with trigonometric equation
Hi all, it's been a while since I took trigonometry in high school and I'm stuck on a current homework problem:

Have to solve for what values of 't' make the equation = 0. 'w' is a constant not equal to 0.
$\displaystyle -w * sin(w * t) + sqrt(3) * w * cos(w * t) = 0$

I simplified it because I figured if 'w' is nonzero then only the inside terms mattered:
$\displaystyle -w*(sin(w*t) - sqrt(3)*cos(w*t)) = 0$

I've tried setting sin(w*t) to cos(w*t + pi/2) and solving but I ran into problems.. If anyone could give me a hint to point me in the right direction that would be awesome, and maybe if you knew of a good trigonometry refresher kicking around on the internet that would be great as well.
• Nov 8th 2009, 09:30 AM
e^(i*pi)
Quote:

Originally Posted by LnVII
Hi all, it's been a while since I took trigonometry in high school and I'm stuck on a current homework problem:

Have to solve for what values of 't' make the equation = 0. 'w' is a constant not equal to 0.
$\displaystyle -w * sin(w * t) + sqrt(3) * w * cos(w * t) = 0$

I simplified it because I figured if 'w' is nonzero then only the inside terms mattered:
$\displaystyle -w*(sin(w*t) - sqrt(3)*cos(w*t)) = 0$

I've tried setting sin(w*t) to cos(w*t + pi/2) and solving but I ran into problems.. If anyone could give me a hint to point me in the right direction that would be awesome, and maybe if you knew of a good trigonometry refresher kicking around on the internet that would be great as well.

$\displaystyle -\omega \,sin(\omega t) + \omega \sqrt3 \,cos(\omega t) = 0$

$\displaystyle \omega sin(\omega t) = \omega \sqrt3 \,cos(\omega t)$

$\displaystyle \omega$ will cancel as it's non-zero. If $\displaystyle t \neq 0$

We can divide through by $\displaystyle cos(\omega t)$

$\displaystyle \frac{sin(\omega t)}{cos (\omega t)} = \sqrt{3}$

Can you solve from there for t in terms of $\displaystyle \omega$?

edit: The graph is attached for $\displaystyle \omega = 2$ and $\displaystyle 0 \leq t \leq 2\pi$
• Nov 8th 2009, 09:41 AM
LnVII
So that would be
$\displaystyle tan(wt) = sqrt(3)$

$\displaystyle w*t = pi/3$

$\displaystyle t = pi/3w$

So 't' would equal pi/3*w? Boy, was I overcomplicating that.. thanks so much for your help!
• Nov 8th 2009, 09:48 AM
e^(i*pi)
Quote:

Originally Posted by LnVII
So that would be
$\displaystyle tan(wt) = sqrt(3)$

$\displaystyle w*t = pi/3$

$\displaystyle t = pi/3w$

So 't' would equal pi/3*w? Boy, was I overcomplicating that.. thanks so much for your help!

That is the main solution but there are others because tan(\omega t) will repeat itself every $\displaystyle \frac{\pi}{\omega}$ radians on the t axis;

$\displaystyle t = \frac{\pi}{3\omega} + k \frac{\pi}{\omega} \: \: , \: k \in \mathbb{Z}$