Giải phương trình \(\frac{1}{x}+\frac{1}{\sqrt{1-x^2}}=2\sqrt{2}\)

ĐK: \(x\in (-1;1)\setminus \left \{ 0 \right \}\)
Đặt \(x=cost \ \ \ \ t\in (0;\pi)\setminus \left \{ \frac{\pi}{2} \right \}\)
Phương trình trở thành \(\sqrt{1-x^2}=\sqrt{1-cos^2t}=\sqrt{sin^2t}=sint\)
\(\frac{1}{cost}+\frac{1}{sint}=2\sqrt{2}\)
\(\Leftrightarrow sint+cost=2\sqrt{2}sint.cost \ (1)\)
Đặt \(a=sint+cost \ \ -\sqrt{2}\leq a\leq \sqrt{2}\)
\(sint.cost=\frac{a^2-1}{2}\)
(1) trở thành \(a=\sqrt{2}(a^2-1)\)
\(\Leftrightarrow \sqrt{2}a^2-a-\sqrt{2}=0 \ \Delta =1+4.2=9,\sqrt{\Delta }=3\)
\(\Bigg \lbrack\begin{matrix} a=\frac{1-3}{2\sqrt{2}}=-\frac{1}{\sqrt{2}}\\ \\ a=\frac{1-3}{2\sqrt{2}}=\sqrt{2} \end{matrix}\)
\(a=-\frac{1}{\sqrt{2}}\)

\(\Leftrightarrow sint+cost=-\frac{1}{\sqrt{2}}\)
\(\Leftrightarrow \sqrt{2}sin(t+\frac{\pi}{4})=-\frac{1}{\sqrt{2}}\)
\(\Leftrightarrow sin(t+\frac{\pi}{4})=-\frac{1}{2}\)
\(\Leftrightarrow \Bigg \lbrack \begin{matrix} t=-\frac{5\pi}{12}+k2\pi\\ \\ t=\frac{11\pi}{12}+k2\pi \end{matrix}\) do \(t\in (0;\pi)\setminus \left \{ \frac{\pi}{2} \right \}\) nên \(t\in \left \{ \frac{11\pi}{12} \right \}\)
\(a=\sqrt{2}\Leftrightarrow sint+cost=\sqrt{2}\)
\(\Leftrightarrow \sqrt{2}sin(t+\frac{\pi}{4})=\sqrt{2}\)
\(\Leftrightarrow sin(t+\frac{\pi}{4})=1\)
\(\Leftrightarrow t+\frac{\pi}{4}=\frac{\pi}{2}+k2\pi\Leftrightarrow t=\frac{\pi}{4}+k2\pi\)
do \(t\in (0;\pi)\setminus \left \{ \frac{\pi}{2} \right \}\) nên \(t\in \left \{ \frac{11\pi}{4} \right \}\)
KL: \(x=cos\frac{11\pi}{12}=cos(\frac{2\pi}{3}+\frac{\pi}{4})=cos\frac{2\pi}{3}.cos\frac{\pi}{4} -sin\frac{2\pi}{3}.sin\frac{\pi}{4})=-\frac{\sqrt{2}}{4}-\frac{\sqrt{6}}{4}\)
\(=-\frac{\sqrt{2}+\sqrt{6}}{4}\)
\(x=\frac{\sqrt{2}}{2}\)