# Thread: [SOLVED] Trig equation - Why can't I get it right?!?

1. ## [SOLVED] Trig equation - Why can't I get it right?!?

Hi. I've been trying to solve this trig. equation 4sin4t + 3cos4t = 0. The answer is meant to be t = $\displaystyle \frac{\pi}{4}-\frac{1}{4}\arctan(\frac{3}{4})$

$\displaystyle 4sin4t + 3cos4t = 0$

$\displaystyle 4sin4t = - 3cos4t$

$\displaystyle tan4t = -\frac{3}{4}$

$\displaystyle 4t = \arctan(\frac{3}{4})$

$\displaystyle t = \frac{1}{4}\arctan(\frac{3}{4})$

2. Originally Posted by UltraGirl
Hi. I've been trying to solve this trig. equation 4sin4t + 3cos4t = 0. The answer is meant to be t = $\displaystyle \frac{\pi}{4}-\frac{1}{4}\arctan(\frac{3}{4})$

$\displaystyle 4sin4t + 3cos4t = 0$

$\displaystyle 4sin4t = - 3cos4t$

$\displaystyle tan4t = -\frac{3}{4}$

$\displaystyle 4t = \arctan({\color{red}-}\frac{3}{4})={\color{red}-}\arctan\frac{3}{4}$
You have dropped the minus-sign, also note that there are infinitely many possibilities, namely

$\displaystyle 4t=-\arctan\frac{3}{4}+n\cdot\pi, n\in\mathbb{Z}$

So by dividing by 4 you get the general solution

$\displaystyle t =n\cdot \frac{\pi}{4}-\arctan\frac{3}{4}$

Now plug in 1 for n and you get the particular solution that you have been given.

3. Originally Posted by UltraGirl
Hi. I've been trying to solve this trig. equation 4sin4t + 3cos4t = 0. The answer is meant to be t = $\displaystyle \frac{\pi}{4}-\frac{1}{4}\arctan(\frac{3}{4})$

$\displaystyle 4sin4t + 3cos4t = 0$

$\displaystyle 4sin4t = - 3cos4t$

$\displaystyle tan4t = -\frac{3}{4}$ specifies a line through the origin with a slope of -3/4

$\displaystyle 4t = \arctan(\frac{3}{4})$ gives you the "acute" angle the line makes with the negative part of the x-axis in the 2nd quadrant

You need to specify the counterclockwise angle the line makes with the positive half of the x-axis

$\displaystyle t = \frac{1}{4}\arctan(\frac{3}{4})$
Hi Ultragirl,

$\displaystyle 4sin4t+3cos4t=0$

$\displaystyle tan4t=-\frac{3}{4}$

$\displaystyle tan4t=tan^{-1}\left(-\frac{3}{4}\right)$

Line through origin, with slope $\displaystyle -\frac{3}{4}$ passes through 2nd and 4th quadrants.

The angle $\displaystyle tan^{-1}\left(\frac{3}{4}\right)$ is the acute angle the line makes with the negative x-axis in the 2nd quadrant
or with the positive x-axis in the 4th quadrant.

The angle 4t is the pair of counterclockwise angles the line with negative slope
makes with the positive x-axis.

$\displaystyle 4t={\pi}-tan^{-1}\left(\frac{3}{4}\right)$

$\displaystyle t=\frac{{\pi}}{4}-\frac{1}{4}tan^{-1}\left(\frac{3}{4}\right)$

For other solutions, add integer multiples of $\displaystyle {\pi}$

In the attachment,

$\displaystyle 4t\ =\ counterclockwise\ angles\ AOB\ and\ AOC$

$\displaystyle \theta=arctan\left(\frac{3}{4}\right)$