Lời giải:Ta có: \(\sin x + \cos x = 1 - \dfrac{1}{2}\sin 2x \) \(\Leftrightarrow \sin x + \cos x = 1 - \sin x\cos x\)
\( \Leftrightarrow \sin x + \cos x - 1 + \sin x\cos x = 0\)
Đặt \(t = \sin x + \cos x\)
\(\begin{array}{l} \Rightarrow {t^2} = {\left( {\sin x + \cos x} \right)^2}\\ = {\sin ^2}x + 2\sin x\cos x + {\cos ^2}x\\ = 1 + 2\sin x\cos x\\ = 1 + \sin 2x \le 1 + 1 = 2\\ \Rightarrow {t^2} \le 2 \Rightarrow - \sqrt 2 \le t \le \sqrt 2 \\ \Rightarrow \sin x\cos x = \frac{{{t^2} - 1}}{2}\end{array}\)
Phương trình trở thành:
\(\begin{array}{l}t - 1 + \frac{{{t^2} - 1}}{2} = 0\\ \Leftrightarrow 2t - 2 + {t^2} - 1 = 0\\ \Leftrightarrow {t^2} + 2t - 3 = 0\\ \Leftrightarrow \left[ \begin{array}{l}t = 1\,\,\,\,\,\,\left( {TM} \right)\\t = - 3\,\,\,\left( {KTM} \right)\end{array} \right.\end{array}\)
Với \(t = 1\) thì \(\sin x + \cos x = 1\)
\(\begin{array}{l} \Leftrightarrow \sqrt 2 \sin \left( {x + \frac{\pi }{4}} \right) = 1\\ \Leftrightarrow \sin \left( {x + \frac{\pi }{4}} \right) = \frac{1}{{\sqrt 2 }} = \sin \frac{\pi }{4}\\ \Leftrightarrow \left[ \begin{array}{l}x + \frac{\pi }{4} = \frac{\pi }{4} + k2\pi \\x + \frac{\pi }{4} = \frac{{3\pi }}{4} + k2\pi \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}x = k2\pi \\x = \frac{\pi }{2} + k2\pi \end{array} \right.\end{array}\)
Chọn đáp án D.
