Đáp án:
Giải thích các bước giải:
4g)
\[\begin{array}{l}
g)\,\,\,\left( {1 + {{\sin }^2}x} \right)\cos x + \left( {1 + {{\cos }^2}x} \right)\sin x = 1 + \sin 2x\\
\Leftrightarrow \cos x + {\sin ^2}x\cos x + \sin x + {\cos ^2}x\sin x = {\sin ^2}x + \cos {x^2} + 2\sin x\cos x\\
\Leftrightarrow \sin x + \cos x + \sin x\cos x\left( {\sin x + \cos x} \right) = {\left( {\sin x + \cos x} \right)^2}\,\,\,\,\left( * \right)\\
Dat\,\,\,\sin x + \cos x = t\,\,\,\left( { - \sqrt 2 \le t \le \sqrt 2 } \right)\\
\Rightarrow {t^2} = {\left( {\sin x + \cos x} \right)^2} \Leftrightarrow {t^2} = 1 + 2\sin x\cos x\\
\Leftrightarrow \sin x\cos x = \frac{{{t^2} - 1}}{2}\\
\Rightarrow \left( * \right) \Leftrightarrow t + \frac{{{t^2} - 1}}{2}.t = {t^2}\\
\Leftrightarrow 2t + {t^3} - t - 2{t^2} = 0\\
\Leftrightarrow {t^3} - 2{t^2} + t = 0\\
\Leftrightarrow t\left( {{t^2} - 2t + 1} \right) = 0\\
\Leftrightarrow t{\left( {t - 1} \right)^2} = 0 \Leftrightarrow \left[ \begin{array}{l}
t = 0\\
t = 1
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
\sin x + \cos x = 0\\
\sin x + \cos x = 1
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
\sqrt 2 \sin \left( {x + \frac{\pi }{4}} \right) = 0\\
\sqrt 2 \sin \left( {x + \frac{\pi }{4}} \right) = 1
\end{array} \right. \Leftrightarrow ......
\end{array}\]
