Phương trình

C

- Hướng dẫn giải

Đáp án C

PT $\left[\begin{array}{l}2x=-\frac{\pi }{4}+k2\pi \\ 2x=\frac{5\pi }{4}+k2\pi \end{array}\right.\iff \left[\begin{array}{l}x=-\frac{\pi }{8}+k\pi \\ x=\frac{5\pi }{8}+k\pi \end{array}\right.\left(k\in \mathbb{Z}\right)$

Vì $x\in \left(0;\pi \right)\Rightarrow \left[\begin{array}{l}0<-\frac{\pi }{8}+k\pi <\pi \\ 0<\frac{5\pi }{8}+k\pi <\pi \end{array}\right.\iff \left[\begin{array}{l}\frac{1}{8}<k<\frac{9}{8}\\ -\frac{5}{8}<k<\frac{3}{8}\end{array}\right.\Rightarrow \left[\begin{array}{l}k=1\\ k=0\end{array}\right.\Rightarrow \left[\begin{array}{l}x=\frac{7\pi }{8}\\ x=\frac{5\pi }{8}\end{array}\right.$