Tan(4x+ 9pi/8) +cot( 2x-3pi/4 )=0 tính tổng các nghiệm trên khoang [ -pi;pi]

Đáp án:

$S = \dfrac{\pi}{4}$

Giải thích các bước giải:

$\begin{array}{l}\tan\left(4x +\dfrac{9\pi}{8}\right) + \cot\left(2x - \dfrac{3\pi}{4}\right) = 0 \,\,\,\, (*)\\ ĐK\begin{cases}\cos\left(4x +\dfrac{9\pi}{8}\right) \ne 0\\
\sin\left(2x - \dfrac{3\pi}{4}\right)\ne 0\end{cases}\Leftrightarrow \begin{cases}x \ne - \dfrac{\pi}{4} + k\dfrac{\pi}{4}\\x \ne \dfrac{3\pi}{4} + k\dfrac{\pi}{2} \end{cases}\,\,\,(k \in \Bbb Z)\\
 (*) \Leftrightarrow \tan\left(\pi + 4x + \dfrac{\pi}{8}\right) = \cot\left(\pi - 2x + \dfrac{3\pi}{4}\right)\\
\Leftrightarrow \tan\left(4x + \dfrac{\pi}{8}\right) = \tan\left(2x - \dfrac{5\pi}{4}\right)\\
\Leftrightarrow 4x + \dfrac{\pi}{8} = 2x - \dfrac{5\pi}{4} + k\pi\\
\Leftrightarrow x = -\dfrac{11\pi}{16} + k\dfrac{\pi}{2}\,\,\,\,\,\,(k \in \Bbb Z)\\
\text{Ta có:}\\ -\pi < x <\pi \\
\Leftrightarrow -\pi < -\dfrac{11\pi}{16} + k\dfrac{\pi}{2} < \pi\\
\Leftrightarrow -\dfrac{5}{8} < k < \dfrac{27}{8}\\
\Rightarrow k = \left\{0;1;2;3\right\}\\
\Rightarrow S = \dfrac{-11\pi}{16}.4 + \dfrac{\pi}{2} + \pi + \dfrac{3\pi}{2} = \dfrac{\pi}{4}\end{array}$