# How do you find the exact values of cos^-1 0?

Set

${\cos}^{-} 1 0 = x$

Hence

$\cos x = 0$

For $0 \le x \le 2 \cdot \pi$ the solutions are $x = \frac{\pi}{2}$ and $x = 3 \cdot \frac{\pi}{2}$

The general solutions are

x = ±π/2 + 2kπ where k is an integer

May 19, 2018

$x = \pm \frac{\pi}{2} + 2 \pi k \quad$ integer $k$

#### Explanation:

It's important to recognize in general the inverse trig functions are multivalued, and that while a particular solution is a necessary step in finding the general solution, it is not in itself the general solution.

The general solution to

$\cos x = \cos a$

is

$x = \pm a + {360}^{\circ} k \quad$ integer $k$

or

$x = \pm a + 2 \pi k \quad$ in radians.

Here we have $x = \arccos 0$ or

$\cos x = 0$

A particular solution is $x = \frac{\pi}{2}$ i.e.,

$\cos x = \cos \left(\frac{\pi}{2}\right)$

Applying our recipe, the general solution is

$x = \pm \frac{\pi}{2} + 2 \pi k$