Câu hỏi 31

\(4\sin x.\cos x + 1 = \cos x - \sin x \Leftrightarrow \sin x - \cos x + 4\sin x.\cos x + 1 = 0\,\,\,\left( * \right)\)

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

\({\left( {\sin x - \cos x} \right)^2} = {t^2} \Leftrightarrow 1 - 2\sin x.\cos x = {t^2} \Leftrightarrow \sin x.\cos x = \dfrac{{1 - {t^2}}}{2}\).

Phương trình trở thành:

\(\begin{array}{l} \Rightarrow (*) \Leftrightarrow t + 4\left( {\dfrac{{1 - {t^2}}}{2}} \right) + 1 = 0 \Leftrightarrow  - 2{t^2} + t + 3 = 0 \Leftrightarrow \left[ \begin{array}{l}t = \dfrac{3}{2}\,\,\,\left( {ktm} \right)\\t =  - 1\,\,\left( {tm} \right)\end{array} \right.\\t =  - 1 \Leftrightarrow \sin x - \cos x =  - 1 \Leftrightarrow \sqrt 2 \sin \left( {x - \dfrac{\pi }{4}} \right) =  - 1\\ \Leftrightarrow \sin \left( {x - \dfrac{\pi }{4}} \right) = \sin \left( {\dfrac{{ - \pi }}{4}} \right) \Leftrightarrow \left[ \begin{array}{l}x - \dfrac{\pi }{4} =  - \dfrac{\pi }{4} + k2\pi \\x - \dfrac{\pi }{4} = \dfrac{{5\pi }}{4} + k2\pi \end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = k2\pi \\x = \dfrac{{3\pi }}{2} + k2\pi \end{array} \right.\,\,\left( {k \in \mathbb{Z}} \right)\end{array}\)

Vậy \(S = \left\{ {k2\pi ;\dfrac{{3\pi }}{2} + k2\pi ,\,\,k \in \mathbb{Z}} \right\}.\)