Câu hỏi 33

\(\sqrt 2 \left( {\sin x + cosx} \right) - 1 = \sin x.cosx\,\,\,\left( * \right)\)

Đặt ,\(\sin x + \cos x = t\,\,\,\,\left( { - \sqrt 2  \le t \le \sqrt 2 } \right)\) khi đó ta có:

\(\begin{array}{l}{\left( {\sin x + cosx} \right)^2} = {t^2} \Leftrightarrow {\sin ^2}x + 2\sin x.\cos x + {\cos ^2}x = {t^2}\\ \Leftrightarrow 1 + 2\sin x.\cos x = {t^2} \Leftrightarrow \sin x.\cos x = \dfrac{{{t^2} - 1}}{2}\end{array}\)
Phương trình trở thành:

\(\begin{array}{l}
\Rightarrow \sqrt 2 t - 1 = \frac{{{t^2} - 1}}{2} \Leftrightarrow {t^2} - 2\sqrt 2 t + 1 = 0 \Leftrightarrow \left[ \begin{array}{l}
t = 1 + \sqrt 2 \,\,\,\,\,\left( {ktm} \right)\\
t = - 1 + \sqrt 2 \,\,\left( {tm} \right)
\end{array} \right.\\
\Leftrightarrow \sin x + \cos x = - 1 + \sqrt 2 \\
\Leftrightarrow \sin \left( {x + \frac{\pi }{4}} \right) = \sin \alpha \,\,\,\,\,\left( {Voi\,\,\sin \alpha = \frac{{ - 1 + \sqrt 2 }}{{\sqrt 2 }}} \right)\\
\Leftrightarrow \left[ \begin{array}{l}
x + \frac{\pi }{4} = \alpha + k2\pi \\
x + \frac{\pi }{4} = \pi - \alpha + k2\pi
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = \alpha - \frac{\pi }{4} + k2\pi \\
x = \frac{{3\pi }}{4} - \alpha + k2\pi
\end{array} \right.\left( {k \in } \right)
\end{array}\)