Đáp án:
$\begin{array}{l}
Đặt:\cos x + \sqrt {1 + {{\sin }^2}x} = t\\
\Rightarrow {\cos ^2}x + 2\cos x.\sqrt {1 + {{\sin }^2}x} + 1 + {\sin ^2}x = {t^2}\\
\Rightarrow 2 + 2\cos x.\sqrt {1 + {{\sin }^2}x} = {t^2}\\
\Rightarrow {\mathop{\rm cosx}\nolimits} .\sqrt {1 + {{\sin }^2}x} = \frac{{{t^2} - 2}}{2}\\
Pt:t + \frac{{{t^2} - 2}}{2} = 3\\
\Rightarrow {t^2} + 2t - 8 = 0\\
\Rightarrow \left( {t - 2} \right)\left( {t + 4} \right) = 0\\
\Rightarrow \left[ \begin{array}{l}
t = 2\\
t = - 4
\end{array} \right. \Rightarrow \left[ \begin{array}{l}
\cos x + \sqrt {1 + {{\sin }^2}x} = 2\\
\cos x + \sqrt {1 + {{\sin }^2}x} = - 4\left( {ktm} \right)
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
\cos x = 1\\
{\sin ^2}x = 0
\end{array} \right.\\
\Rightarrow x = k2\pi \left( {k \in Z} \right)
\end{array}$
