Đáp án:

\(x = \frac{\pi }{2}\)

Giải thích các bước giải: \[\begin{array}{l} \sin 2x + 4\left( {\cos x - \sin x} \right) = 4\,\,\,\left( * \right)\\ Dat\,\,\,t = \cos x - \sin x\,\,\,\left( { - \sqrt 2 \le t \le \sqrt 2 } \right)\\ \Rightarrow {t^2} = 1 - 2\sin x\cos x = 1 - \sin 2x\\ \Rightarrow \sin 2x = 1 - {t^2}\\ \Rightarrow \left( * \right) \Leftrightarrow 1 - {t^2} + 4t = 4\\ \Leftrightarrow {t^2} - 4t + 3 = 0 \Leftrightarrow \left[ \begin{array}{l} t = 1\,\,\,\left( {tm} \right)\\ t = 3\,\,\,\left( {ktm} \right) \end{array} \right.\\ \Leftrightarrow \cos x - \sin x = 1\\ \Leftrightarrow \sqrt 2 \cos \left( {x - \frac{\pi }{4}} \right) = 1\\ \Leftrightarrow \cos \left( {x - \frac{\pi }{4}} \right) = \frac{1}{{\sqrt 2 }}\\ \Leftrightarrow \left[ \begin{array}{l} x - \frac{\pi }{4} = \frac{\pi }{4} + k2\pi \\ x - \frac{\pi }{4} = - \frac{\pi }{4} + m2\pi \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x = \frac{\pi }{2} + k2\pi \\ x = m2\pi \end{array} \right.\,\,\,\left( {k,\,\,m \in Z} \right).\\ x \in \left( {0;\,\,2\pi } \right) \Rightarrow \left[ \begin{array}{l} 0 < \frac{\pi }{2} + k2\pi < 2\pi \\ 0 < m2\pi < 2\pi \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l} - \frac{\pi }{2} < k2\pi < \frac{{3\pi }}{2}\\ 0 < m < 1 \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} - \frac{1}{4} < k < \frac{3}{4}\\ m \in \emptyset \end{array} \right.\\ \Leftrightarrow k = 0\\ \Rightarrow x = \frac{\pi }{2}. \end{array}\]