For y = f(x), the domain of the function f(x) is all the values of x for which y is real.

Here y = f(x) = sqrt[cos(x)]

The value of cos x lies in the interval [-1, 1] for all values of x.

But sqrt [cos(x)] is real only when cos(x) is not negative. Also it has to be kept in mind that cos(x) is a periodic function and we get the same value for cos x after x has decreased on increased by 2*pi.

The interval for x where the value of cos x is not negative is [0 + 2*n*pi, pi/2 + 2*n*pi] U [3*pi/2 + 2*n*pi, 0 + 2*n*pi]

**The domain of the function f(x) = sqrt[cos(x)] is [0 + 2*n*pi, pi/2 + 2*n*pi] U [3*pi/2 + 2*n*pi, 0 + 2*n*pi]**

First, we'll impose the condition for the square root to exist.

cos x >= 0

We know that the cosine function is positive if the angle x belongs to the 1st or the 4th quadrants.

The 1st quadrant covers the interval [0 ; pi/2].

The 4th quadrant covers the interval [3pi/2 ; 2pi].

**Therefore, the domain of the function f(x) = sqrt(cos x) is the reunion of intervals [0 ; pi/2]U[3pi/2 ; 2pi].**