# Math Help - Help with Solving Trigonometric Equations

1. ## Help with Solving Trigonometric Equations

Okay, so I have the function,
$\cos x = -\frac{1}{\sqrt{2}}$ for $-\pi \le x \le \pi$

Basic angle $= \cos^{-1}\left(\frac{1}{\sqrt{2}}\right)=\frac{\pi}{4}$

$\cos$ is negative in 2nd + 3rd quadrants

$x= \cos^{-1}\left(\frac{1}{\sqrt{2}}\right)$

$x= \pi-\frac{\pi}{4}, \pi + \frac{\pi}{4}$

$x= \frac{3\pi}{4}, \frac{5\pi}{4}$

$\frac{3\pi}{4}$ is one solution. However $\frac{5\pi}{4}$ is not in the domain so I assume it it not a solution?

The book tells me there is another solution $= -\frac{3\pi}{4}$

How do I obtain this solution?

It is confusing when going in a clockwise direction of the unit circle.

Thanks.

2. Originally Posted by user_5
Okay, so I have the function,
$\cos x = -\frac{1}{\sqrt{2}}$ for $-\pi \le x \le \pi$

Basic angle $= \cos^{-1}\left(\frac{1}{\sqrt{2}}\right)=\frac{\pi}{4}$

$\cos$ is negative in 2nd + 3rd quadrants

$x= \cos^{-1}\left(\frac{1}{\sqrt{2}}\right)$

$x= \pi-\frac{\pi}{4}, \pi + \frac{\pi}{4}$

$x= \frac{3\pi}{4}, \frac{5\pi}{4}$

$\frac{3\pi}{4}$ is one solution. However $\frac{5\pi}{4}$ is not in the domain so I assume it it not a solution?

The book tells me there is another solution $= -\frac{3\pi}{4}$

How do I obtain this solution?

It is confusing when going in a clockwise direction of the unit circle.

Thanks.
Remember that the cosine function is an even function, and therefore, if your interval is symmetrical, then whatever solutions you get in the positive side of the interval also exist in the negative side of the interval.

3. Okay, is there a way I can show working out for the solution $
-\frac{3\pi}{4}$
?

Like how $\frac{3\pi}{4}$ is found from $\pi-\frac{\pi}{4}$

4. Originally Posted by user_5
Okay, is there a way I can show working out for the solution $
-\frac{3\pi}{4}$
?

Like how $\frac{3\pi}{4}$ is found from $\pi-\frac{\pi}{4}$

Sure. When we call a function 'even' we can write this mathematically as:

$\cos(x) = \cos(-x)$, you can work from here to show that any solution that results in considering $\cos(x)$ should also find you a symmetrical solution by considering $\cos(-x)$.