Đáp án:
$A.\,\dfrac{4071315\pi}{2}$
Giải thích các bước giải:
$\sin2x = 1$
$\Leftrightarrow x = \dfrac{\pi}{4} + k\pi\quad (k\in\Bbb Z)$
Ta có: $0 \leq x \leq 2018\pi$
$\Leftrightarrow 0\leq \dfrac{\pi}{4} + k\pi\leq 2018\pi$
$\Leftrightarrow -0,25 \leq k \leq 2017,75$
Do $k\in \Bbb Z$
nên $k = \left\{0;1;2;3;\dots;2016;0217\right\}$
$\Rightarrow \mathop{\sum}\limits_{k = 0}^{2017}\dfrac{\pi}{4} + k\pi = \dfrac{\pi}{4} .2018 + \dfrac{2017.(2017 + 1)}{2}\pi = \dfrac{4071315\pi}{2}$
