1 câu trả lời
Đáp án:
\(\left[ \begin{array}{l} x = - \frac{\pi }{2} + k2\pi \\ x = \pi + k2\pi \end{array} \right.\,\,\,\left( {k \in Z} \right)\)
Giải thích các bước giải: \[\begin{array}{l} 5\left( {\sin x + \cos x + 1} \right) = 12\sin 2x\\ \Leftrightarrow 5\left( {\sin x + \cos x} \right) - 24\sin x\cos x + 5 = 0\,\,\,\,\left( 1 \right)\\ Dat\,\,\,\sin x + \cos x = t\,\,\,\left( { - \sqrt 2 \le t \Leftrightarrow \sqrt 2 } \right)\\ \Rightarrow {t^2} = 1 + 2\sin x\cos x \Rightarrow 2\sin x\cos x = {t^2} - 1\\ \Rightarrow \left( 1 \right) \Leftrightarrow 5t - 12\left( {{t^2} - 1} \right) + 5 = 0\\ \Leftrightarrow 5t - 12{t^2} + 12 + 5 = 0\\ \Leftrightarrow 12{t^2} - 5t - 17 = 0 \Leftrightarrow \left[ \begin{array}{l} t = \frac{{17}}{{12}}\,\,\,\left( {ktm} \right)\\ t = - 1\,\,\,\left( {tm} \right) \end{array} \right.\\ \Rightarrow \sin x + \cos x = - 1\\ \Leftrightarrow \sqrt 2 \sin \left( {x + \frac{\pi }{4}} \right) = - 1\\ \Leftrightarrow \sin \left( {x + \frac{\pi }{4}} \right) = - \frac{1}{{\sqrt 2 }}\\ \Leftrightarrow \left[ \begin{array}{l} x + \frac{\pi }{4} = - \frac{\pi }{4} + k2\pi \\ x + \frac{\pi }{4} = \frac{{5\pi }}{4} + k2\pi \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x = - \frac{\pi }{2} + k2\pi \\ x = \pi + k2\pi \end{array} \right.\,\,\,\left( {k \in Z} \right). \end{array}\]
