Giải thích các bước giải:
ĐKXĐ: \(x \ne \frac{{k\pi }}{2}\)
Ta có:
\(\begin{array}{l}
\frac{3}{{{{\sin }^2}x}} + {\tan ^2}x + \tan x + \cot x = 1\\
\Leftrightarrow \frac{3}{{{{\sin }^2}x}} + \frac{{{{\sin }^2}x}}{{{{\cos }^2}x}} + \frac{{\sin x}}{{\cos x}} + \frac{{\cos x}}{{\sin x}} = 1\\
\Leftrightarrow \frac{3}{{{{\sin }^2}x}} + \frac{{{{\sin }^2}x}}{{{{\cos }^2}x}} + \frac{{{{\sin }^2}x + {{\cos }^2}x}}{{\sin x.\cos x}} = 1\\
\Leftrightarrow \frac{3}{{{{\sin }^2}x}} + \frac{{{{\sin }^2}x}}{{{{\cos }^2}x}} + \frac{1}{{\sin x.\cos x}} = 1\\
\Leftrightarrow \frac{{3{{\cos }^2}x + {{\sin }^4}x + \sin x\cos x}}{{{{\sin }^2}x{{\cos }^2}x}} = 1\\
\Leftrightarrow 3{\cos ^2}x + {\sin ^4}x + \sin x\cos x = {\sin ^2}x{\cos ^2}x\\
\Leftrightarrow {\sin ^4}x + 3.\left( {1 - {{\sin }^2}x} \right) = {\left( {\sin x\cos x} \right)^2} - \sin x\cos x\\
\Leftrightarrow {\sin ^4}x - 3{\sin ^2}x + 3 = {\left( {\sin x\cos x} \right)^2} - \sin x\cos x\\
\Leftrightarrow {\left( {{{\sin }^2}x - \frac{3}{2}} \right)^2} + 1 = {\left( {\sin x\cos x - \frac{1}{2}} \right)^2}\,\,\,\,\left( 1 \right)\\
{\sin ^2}x.{\cos ^2}x = {\sin ^2}x\left( {1 - {{\sin }^2}x} \right) = - {\sin ^4}x + {\sin ^2}x = \frac{1}{4} - {\left( {{{\sin }^2}x - \frac{1}{2}} \right)^2} \le \frac{1}{4}\\
\Rightarrow - \frac{1}{2} \le \sin x\cos x \le \frac{1}{2}\\
\Rightarrow - 1 \le \sin x\cos x - \frac{1}{2} \le 0\\
\Leftrightarrow 0 \le {\left( {\sin x\cos x - \frac{1}{2}} \right)^2} \le 1\\
{\left( {{{\sin }^2}x - \frac{3}{2}} \right)^2} + 1 \ge 1 \ge {\left( {\sin x\cos x - \frac{1}{2}} \right)^2}\,\,\,\,\left( 2 \right)\\
\left( 1 \right);\left( 2 \right) \Rightarrow \left\{ \begin{array}{l}
{\sin ^2}x - \frac{3}{2} = 0\\
\sin x\cos x = - \frac{1}{2}
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
\sin x = \frac{{ \pm \sqrt 6 }}{2}\\
\sin x\cos x = - \frac{1}{2}
\end{array} \right.\left( {vn} \right)
\end{array}\)
Vậy phương trình đã cho vô nghiệm
