Find the sum of the smallest positive and largest negative roots of the equation cosx = √3 / 2.

1. Let’s find the general solution of this trigonometric equation:

cosx = √3 / 2;
x = ± π / 6 + 2πk, k ∈ Z;
x1 = -π / 6 + 2πk, k ∈ Z;
x2 = π / 6 + 2πk, k ∈ Z.
2. The largest negative and smallest positive roots of the equation correspond to the values of x1 and x2 at k = 0:

x1 = -π / 6 + 2π * 0 = -π / 6;
x2 = π / 6 + 2π * 0 = π / 6.
3. The sum of these roots is zero:

x1 + x2 = -π / 6 + π / 6 = 0.

The answer is zero.