Câu hỏi 29

\(\left| {\cos x - \sin x} \right| + 6\sin x\cos x = 1. \Leftrightarrow \left| {\sin x - \cos x} \right| + 6\sin x.\cos x = 1\,\,\,\left( * \right)\)

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: \(\left| t \right| + 6\left( {\dfrac{{1 - {t^2}}}{2}} \right) = 1 \Leftrightarrow \left| t \right| + 3 - 3{t^2} = 1\,\,\,\left( 1 \right)\)

TH1: \(0 \le t \le \sqrt 2 \).

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

TH2: \( - \sqrt 2  \le t < 0\).

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

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