Giải phương trình $\sqrt {\frac{{1 + \sin x}}{{1 - \sin x}}} + \sqrt {\frac{{1 - \sin x}}{{1 + \sin x}}} = \frac{4}{{\sqrt 3 }}$ với $x \in \left( {0;\frac{\pi }{2}} \right)$.
A. $x = \frac{\pi }{{12}}$.
B. $x = \frac{\pi }{4}$.
C. $x = \frac{\pi }{3}$.
D. $x = \frac{\pi }{6}$.
A. $x = \frac{\pi }{{12}}$.
B. $x = \frac{\pi }{4}$.
C. $x = \frac{\pi }{3}$.
D. $x = \frac{\pi }{6}$.
Chọn A.
${\rm{pt}} \Leftrightarrow \frac{{1 + \sin x + 1 - \sin x}}{{\sqrt {1 - {{\sin }^2}x} }} = \frac{4}{{\sqrt 3 }} \Leftrightarrow \frac{2}{{\cos x}} = \frac{4}{{\sqrt 3 }} \Leftrightarrow \cos x = \frac{{\sqrt 3 }}{2} \Leftrightarrow x = \pm \frac{\pi }{{12}} + k\pi $.
Do $x \in \left( {0;\frac{\pi }{2}} \right)$ nên $x = \frac{\pi }{{12}}$.
${\rm{pt}} \Leftrightarrow \frac{{1 + \sin x + 1 - \sin x}}{{\sqrt {1 - {{\sin }^2}x} }} = \frac{4}{{\sqrt 3 }} \Leftrightarrow \frac{2}{{\cos x}} = \frac{4}{{\sqrt 3 }} \Leftrightarrow \cos x = \frac{{\sqrt 3 }}{2} \Leftrightarrow x = \pm \frac{\pi }{{12}} + k\pi $.
Do $x \in \left( {0;\frac{\pi }{2}} \right)$ nên $x = \frac{\pi }{{12}}$.