If the sum of inverse cosines sum_(i=1)²ⁿ cos⁻¹(x_i) = 0, find the value of sum_(i=1)²ⁿ x_i.
0
n
2n
4n
Correct answer: 2n
Solution
The range of cos⁻¹ is [0, pi], so each cos⁻¹(x_i) >= 0. A sum of 2n nonnegative numbers equals 0 only if each is 0. cos⁻¹(x_i) = 0 means x_i = cos(0) = 1. There are 2n such terms, so sum x_i = 2n*1 = 2n.