[SOLVED] Trig equation

**UltraGirl** · May 4th 2010, 10:34 AM

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

$ 4sin4t + 3cos4t = 0 $

$ 4sin4t = - 3cos4t $

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

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

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

**Failure** · May 4th 2010, 11:21 AM

Originally Posted by UltraGirl

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

$ 4sin4t + 3cos4t = 0 $

$ 4sin4t = - 3cos4t $

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

$ 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

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

So by dividing by 4 you get the general solution

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

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

**Archie Meade** · May 4th 2010, 12:27 PM

Originally Posted by UltraGirl

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

$ 4sin4t + 3cos4t = 0 $

$ 4sin4t = - 3cos4t $

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

$ 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

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

Hi Ultragirl,

$4sin4t+3cos4t=0$

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

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

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

The angle $tan^{-1}\left(\frac{3}{4}\right)$ 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.

In the 2nd quadrant...

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

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

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

In the attachment,

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

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

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