Question

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

Answer 1

Set

`cos^-1 0=x`

Hence

`cosx=0`

For `0<=x<=2*pi` the solutions are `x=pi/2` and `x=3*pi/2`

The general solutions are

`x = ±π/2 + 2kπ` where k is an integer

Answer 2

`x = pm pi/2 + 2pi 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 + 2pi k quad` in radians.

Here we have `x = arccos 0` or

`cos x = 0`

A particular solution is `x=pi/2` i.e.,

`cos x = cos (pi/2)`

Applying our recipe, the general solution is

`x = pm pi/2 + 2pi k`