Find the least value of $k$ for which $f(x) = kx - 2\sin x$ is increasing on $\mathbb{R}$.

Explanation

1. 01

   $f'(x) = k - 2\cos x \ge 0$ for every $x$.

2. 02

   The largest value of $2\cos x$ is 2, so $k \ge 2$; the least $k$ is 2.