6$\sqrt[]{2}$sin(x+45)-5sin2x-3=0. Giải nghiệm giúp

Đáp án:

\(\left[ \begin{array}{l} x = \alpha - \frac{\pi }{4} + k2\pi \\ x = \frac{{3\pi }}{4} - \alpha + k2\pi \end{array} \right.\,\,\left( {k \in Z} \right)\)

Giải thích các bước giải: \(\begin{array}{l} 6\sqrt 2 \sin \left( {x + {{45}^0}} \right) - 5\sin 2x - 3 = 0\\ \Leftrightarrow 6\left( {\sin x + \cos x} \right) - 5\sin 2x - 3 = 0\,\,\,\,\left( * \right)\\ Dat\,\,\,\sin x + \cos x = t\,\,\left( { - \sqrt 2 \le t \le \sqrt 2 } \right)\\ \Rightarrow {t^2} = 1 + 2\sin x\cos x = 1 + \sin 2x\\ \Rightarrow \sin 2x = {t^2} - 1\\ \Rightarrow \left( * \right) \Leftrightarrow 6t - 5\left( {{t^2} - 1} \right) - 3 = 0\\ \Leftrightarrow 6t - 5{t^2} + 5 - 3 = 0\\ \Leftrightarrow 5{t^2} - 6t - 2 = 0\\ \Leftrightarrow \left[ \begin{array}{l} t = \frac{{3 + \sqrt {19} }}{5}\,\,\,\left( {ktm} \right)\\ t = \frac{{3 - \sqrt {19} }}{5}\,\,\left( {tm} \right) \end{array} \right.\\ \Rightarrow \sin x + \cos x = \frac{{3 - \sqrt {19} }}{5}\\ \Leftrightarrow \sqrt 2 \sin \left( {x + \frac{\pi }{4}} \right) = \frac{{3 - \sqrt {19} }}{5}\\ \Leftrightarrow \sin \left( {x + \frac{\pi }{4}} \right) = \frac{{3 - \sqrt {19} }}{{5\sqrt 2 }} = \sin \alpha \\ \Leftrightarrow \left[ \begin{array}{l} x + \frac{\pi }{4} = \alpha + k2\pi \\ x + \frac{\pi }{4} = \pi - \alpha + k2\pi \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x = \alpha - \frac{\pi }{4} + k2\pi \\ x = \frac{{3\pi }}{4} - \alpha + k2\pi \end{array} \right.\,\,\left( {k \in Z} \right). \end{array}\)