What is the domain of #f(x) = sqrt(cosx)#?

1 Answer
Feb 11, 2018

The domain is #-pi/2 + 2pin ≤ x ≤ pi/2 + 2pin#

Explanation:

We have

#f(x) = sqrt(cosx)#

We know that #sqrt#s can never have a negative value under them.

Thus, we need

#cosx ≥ 0#

We solve this inequality by converting it into an equation then using test points.

#cosx = 0#

#x = pi/2, (3pi)/2, ...#

Now if we test #x = pi#, we see that #cos(pi) = -1#, which is smaller than #0#, therefore the inequality is not satisfied and #(pi/2, (3pi)/2)# will not be within the domain of #f(x)#.

However, when #x = 0#, we will be within the domain, because #cos(0) = 1#, and #1 > 0#. Thus our solution set here will be #[-pi/2, pi/2]#. This will restart at #(3pi)/2#. Therefore our general solution can be #-pi/2 + 2pin ≤ x ≤ pi/2 + 2pin#.

Hopefully this helps!