Trigonometry Summation

For $\displaystyle \sum_{k=1}^{100} \left(\sin x_k + \frac 1{\sin x_k}\right) = 200$ , $\sin x_k > 0$ and we can apply AM-GM inequality as: $\sin x_k + \dfrac 1{\sin x_k} \ge 2 \sqrt{\dfrac {\sin x_k}{\sin x_k}} = 2$ . And equality occurs when $\sin x_k = 1$ or $x_k = \dfrac \pi 2$ for $x_k \in (0, \pi)$ for all $k$ . Then $\displaystyle \sum_{k=1}^{100} \left(\sin x_k + \frac 1{\sin x_k}\right) = 200$ has only one solution that is $x_k = \dfrac \pi 2$ for all $k$ . Therefore, $\displaystyle \sum_{k=1}^{100} \cos^k x_k = \sum_{k=1}^{100} \cos^k \frac \pi 2 = \boxed 0$ .