Các nghiệm thuộc khoảng $\left( {0;\pi } \right)$ của phương trình: $\sqrt {\tan x + \sin x} + \sqrt {\tan x - \sin x} = \sqrt {3\tan x} $ là:
A. $\frac{\pi }{8},\frac{{5\pi }}{8}$.
B. $\frac{\pi }{4},\frac{{3\pi }}{4}$.
C. $\frac{\pi }{6},\frac{{5\pi }}{6}$.
D. $\frac{\pi }{6}$.
A. $\frac{\pi }{8},\frac{{5\pi }}{8}$.
B. $\frac{\pi }{4},\frac{{3\pi }}{4}$.
C. $\frac{\pi }{6},\frac{{5\pi }}{6}$.
D. $\frac{\pi }{6}$.
Chọn D
$\sqrt {\tan x + \sin x} + \sqrt {\tan x - \sin x} = \sqrt {3\tan x} $
$ \Rightarrow 2\tan x + 2\sqrt {{{\tan }^2}x - {{\sin }^2}x} = 3\tan x$
$ \Rightarrow 2\sqrt {{{\sin }^2}x\left( {\frac{1}{{{{\cos }^2}x}} - 1} \right)} = \tan x$$ \Rightarrow 2\sqrt {{{\sin }^2}x.{{\tan }^2}x} = \tan x$$ \Rightarrow 4{\sin ^2}x.{\tan ^2}x = {\tan ^2}x$
$ \Rightarrow \left[ \begin{array}{l}{\tan ^2}x = 0\\4{\sin ^2}x = 1\end{array} \right.$$ \Rightarrow \left[ \begin{array}{l}x = k\pi \\\cos 2x = \frac{1}{2}\end{array} \right.$$ \Rightarrow \left[ \begin{array}{l}x = k\pi \\2x = \pm \frac{\pi }{3} + k2\pi \end{array} \right. \Rightarrow \left[ \begin{array}{l}x = k\pi \\x = \pm \frac{\pi }{6} + k\pi \end{array} \right.$
$x \in \left( {0;\pi } \right) \Rightarrow x = \frac{\pi }{6},x = \frac{{5\pi }}{6}$
Thử lại, ta nhận $x = \frac{\pi }{6}$. (Tại $x = \frac{{5\pi }}{6}$ thì $\tan x - \sin x < 0$)
$\sqrt {\tan x + \sin x} + \sqrt {\tan x - \sin x} = \sqrt {3\tan x} $
$ \Rightarrow 2\tan x + 2\sqrt {{{\tan }^2}x - {{\sin }^2}x} = 3\tan x$
$ \Rightarrow 2\sqrt {{{\sin }^2}x\left( {\frac{1}{{{{\cos }^2}x}} - 1} \right)} = \tan x$$ \Rightarrow 2\sqrt {{{\sin }^2}x.{{\tan }^2}x} = \tan x$$ \Rightarrow 4{\sin ^2}x.{\tan ^2}x = {\tan ^2}x$
$ \Rightarrow \left[ \begin{array}{l}{\tan ^2}x = 0\\4{\sin ^2}x = 1\end{array} \right.$$ \Rightarrow \left[ \begin{array}{l}x = k\pi \\\cos 2x = \frac{1}{2}\end{array} \right.$$ \Rightarrow \left[ \begin{array}{l}x = k\pi \\2x = \pm \frac{\pi }{3} + k2\pi \end{array} \right. \Rightarrow \left[ \begin{array}{l}x = k\pi \\x = \pm \frac{\pi }{6} + k\pi \end{array} \right.$
$x \in \left( {0;\pi } \right) \Rightarrow x = \frac{\pi }{6},x = \frac{{5\pi }}{6}$
Thử lại, ta nhận $x = \frac{\pi }{6}$. (Tại $x = \frac{{5\pi }}{6}$ thì $\tan x - \sin x < 0$)