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 $k = \left\{0;1;2;3;\dots;2016;0217\right\}$

$\Rightarrow \mathop{\sum}\limits_{k = 0}^{2017}\dfrac{\pi}{4} + k\pi = \dfrac{\pi}{4} .2018 + \dfrac{2017.(2017 + 1)}{2}\pi = \dfrac{4071315\pi}{2}$

Đáp án: A

Giải thích các bước giải:

$\sin2x=1$

$\Leftrightarrow 2x=\dfrac{\pi}{2}+k2\pi$

$\Leftrightarrow x=\dfrac{\pi}{4}+k\pi$

$0\le x\le 2018\pi$

$\Rightarrow 0\le \dfrac{\pi}{4}+k\pi\le 2018\pi$

$\Leftrightarrow -0,25\le k\le 2017,75$

$\Rightarrow k\in\{0;1;2;...;2017\}$ (2018 số k)

$\Rightarrow x\in\{\dfrac{\pi}{4};\dfrac{5\pi}{4};...;\dfrac{8069\pi}{4}\}$

Tổng nghiệm:

$(\dfrac{\pi}{4}+\dfrac{8069\pi}{4}).2018:2=\dfrac{4071315\pi}{2}$