Đáp án:
$2018$ nghiệm
Giải thích các bước giải:
$\quad F(x)=\displaystyle\int f(x)dx$
$\to F(x)= \displaystyle\int\dfrac{1-\sin^3x}{\sin^2x}dx$
$\to F(x)= \displaystyle\int\dfrac{dx}{\sin^2x} -\displaystyle\int\sin xdx$
$\to F(x)= -\cot x +\cos x + C$
Ta lại có:
$\quad F\left(\dfrac{\pi}{4}\right)= \dfrac{\sqrt2}{2}$
$\to -\cot\dfrac{\pi}{4} +\cos\dfrac{\pi}{4} + C =\dfrac{\sqrt2}{2}$
$\to C = 1$
Ta được:
$\quad F(x)= -\cot x + \cos x + 1$
Ta có:
$\quad F(x)= 1$
$\to \cos x -\cot x = 0$
$\to \sin x\cos x -\cos x = 0$
$\to \cos x(\sin x -1)=0$
$\to \left[\begin{array}{l}\cos x = 0\\\sin x = 1\end{array}\right.$
$\to \left[\begin{array}{l}x =\dfrac{\pi}{2} + \pi\\x =\dfrac{\pi}{2} + k2\pi\end{array}\right.$
$\to x = \dfrac{\pi}{2} + k\pi\quad (k\in\Bbb Z)$
Mặt khác: $0< x < 2018\pi$
$\to 0 < \dfrac{\pi}{2} + k\pi< 2018\pi$
$\to -\dfrac12 < k < 2018 -\dfrac12$
$\to k \in \underbrace{\{0;1;2;\dots;2016;2017\}}_{\text{2018 giá trị k}}$
$\to 2018$ nghiệm
