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

2 Answers

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

May 19, 2018

# 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 #