Nghiệm của phương trình $3\tan \frac{x}{4} - \sqrt 3 = 0$ trong nửa khoảng $\left[ {0;2\pi } \right)$ là
A. $\left\{ {\frac{\pi }{3};\frac{{2\pi }}{3}} \right\}$.
B. $\left\{ {\frac{{3\pi }}{2}} \right\}$.
C. $\left\{ {\frac{\pi }{2};\frac{{3\pi }}{2}} \right\}$.
D. $\left\{ {\frac{{2\pi }}{3}} \right\}$.
A. $\left\{ {\frac{\pi }{3};\frac{{2\pi }}{3}} \right\}$.
B. $\left\{ {\frac{{3\pi }}{2}} \right\}$.
C. $\left\{ {\frac{\pi }{2};\frac{{3\pi }}{2}} \right\}$.
D. $\left\{ {\frac{{2\pi }}{3}} \right\}$.
Chọn D.
$3\tan \frac{x}{4} - \sqrt 3 = 0 \Leftrightarrow \tan \frac{x}{4} = \frac{{\sqrt 3 }}{3} \Leftrightarrow \frac{x}{4} = \frac{\pi }{6} + k\pi \Leftrightarrow x = \frac{{2\pi }}{3} + k4\pi ,k \in \mathbb{Z}$.
Mà $x \in \left[ {0;2\pi } \right) \Leftrightarrow x = \frac{{2\pi }}{3}$.
$3\tan \frac{x}{4} - \sqrt 3 = 0 \Leftrightarrow \tan \frac{x}{4} = \frac{{\sqrt 3 }}{3} \Leftrightarrow \frac{x}{4} = \frac{\pi }{6} + k\pi \Leftrightarrow x = \frac{{2\pi }}{3} + k4\pi ,k \in \mathbb{Z}$.
Mà $x \in \left[ {0;2\pi } \right) \Leftrightarrow x = \frac{{2\pi }}{3}$.