Giải phương trình Cosx - √3 sinx =√2 3sin3x - 4cos3x = 5 2sinx + 2 cosx -√2 =0 5cos2x + 12sin2x - 13=0

Đáp án:

\[\begin{array}{l} a)\,\,\,\left[ \begin{array}{l} x = - \frac{\pi }{{12}} + k2\pi \\ x = - \frac{{7\pi }}{{12}} + k2\pi \end{array} \right.\,\,\,\left( {k \in Z} \right)\\ b)\,\,\,x = \frac{\alpha }{3} + \frac{\pi }{6} + \frac{{k2\pi }}{3}\,\,\,\left( {k \in Z} \right)\\ c)\,\,\left[ \begin{array}{l} x = - \frac{\pi }{{12}} + k2\pi \\ x = \frac{{7\pi }}{{12}} + k2\pi \end{array} \right.\,\,\,\left( {k \in Z} \right)\\ d)\,\,\,x = \frac{\alpha }{2} + k\pi \,\,\,\left( {k \in Z} \right) \end{array}\]

Giải thích các bước giải: \[\begin{array}{l} 1)\,\,\,\cos x - \sqrt 3 \sin x = \sqrt 2 \\ \Leftrightarrow \frac{1}{2}\cos x - \frac{{\sqrt 3 }}{2}\sin x = \frac{{\sqrt 2 }}{2}\\ \Leftrightarrow \cos \left( {x + \frac{\pi }{3}} \right) = \cos \frac{\pi }{4}\\ \Leftrightarrow \left[ \begin{array}{l} x + \frac{\pi }{3} = \frac{\pi }{4} + k2\pi \\ x + \frac{\pi }{3} = - \frac{\pi }{4} + k2\pi \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x = - \frac{\pi }{{12}} + k2\pi \\ x = - \frac{{7\pi }}{{12}} + k2\pi \end{array} \right.\,\,\,\left( {k \in Z} \right)\\ 2)\,\,\,3\sin 3x - 4\cos 3x = 5\\ \Leftrightarrow \frac{3}{5}\sin 3x - \frac{4}{5}\cos 3x = 1\\ \Leftrightarrow \sin \left( {3x - \alpha } \right) = 1\,\,\,\,\left( {voi\,\,\,\,\cos \alpha = \frac{3}{5};\,\,\,\sin \alpha = \frac{4}{5}} \right)\\ \Leftrightarrow 3x - \alpha = \frac{\pi }{2} + k2\pi \\ \Leftrightarrow x = \frac{\alpha }{3} + \frac{\pi }{6} + \frac{{k2\pi }}{3}\,\,\,\left( {k \in Z} \right)\\ 3)\,\,\,2\sin x + 2\cos x - \sqrt 2 = 0\\ \Leftrightarrow \sin x + \cos x = \frac{{\sqrt 2 }}{2}\\ \Leftrightarrow \sqrt 2 \sin \left( {x + \frac{\pi }{4}} \right) = \frac{{\sqrt 2 }}{2}\\ \Leftrightarrow \sin \left( {x + \frac{\pi }{4}} \right) = \frac{1}{2}\, = \sin \frac{\pi }{6}\\ \Leftrightarrow \left[ \begin{array}{l} x + \frac{\pi }{4} = \frac{\pi }{6} + k2\pi \\ x + \frac{\pi }{4} = \pi - \frac{\pi }{6} + k2\pi \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x = - \frac{\pi }{{12}} + k2\pi \\ x = \frac{{7\pi }}{{12}} + k2\pi \end{array} \right.\,\,\,\left( {k \in Z} \right)\\ 4)\,\,\,5\cos 2x + 12\sin 2x - 13 = 0\\ \Leftrightarrow \frac{5}{{13}}\cos 2x + \frac{{12}}{{13}}\sin 2x = 1\\ \Leftrightarrow \cos \left( {2x - \alpha } \right) = 1\,\,\,\left( {voi\,\,\,\cos \alpha = \frac{5}{{13}};\,\,\,\sin \alpha = \frac{{12}}{{13}}} \right)\\ \Leftrightarrow 2x - \alpha = k2\pi \\ \Leftrightarrow 2x = \alpha + k2\pi \\ \Leftrightarrow x = \frac{\alpha }{2} + k\pi \,\,\,\left( {k \in Z} \right). \end{array}\]