Giải các phương trình sau

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Giải các phương trình sau

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12 months 2020-11-08T00:20:16+00:00 1 Answers 44 views 0

$$\begin{array}{l} 15,\\ \sin 2x = 2\cos x + \sqrt 3 .\left( {\sin x – 1} \right)\\ \Leftrightarrow 2\sin x.\cos x = 2\cos x + \sqrt 3 \left( {\sin x – 1} \right)\\ \Leftrightarrow \left( {2\sin x.\cos x – 2\cos x} \right) – \sqrt 3 \left( {\sin x – 1} \right) = 0\\ \Leftrightarrow 2\cos x.\left( {\sin x – 1} \right) – \sqrt 3 \left( {\sin x – 1} \right) = 0\\ \Leftrightarrow \left( {\sin x – 1} \right)\left( {2\cos x – \sqrt 3 } \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l} \sin x = 1\\ \cos x = \dfrac{{\sqrt 3 }}{2} \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x = \dfrac{\pi }{2} + k2\pi \\ x = \pm \dfrac{\pi }{6} + k2\pi \end{array} \right.\\ 16,\\ 2{\sin ^2}\left( {\dfrac{\pi }{4} – 2x} \right) + \sqrt 3 \cos 4x = 4{\cos ^2}x – 1\\ \Leftrightarrow 2.{\left( {\sin \dfrac{\pi }{4}.\cos 2x – \cos \dfrac{\pi }{4}.\sin 2x} \right)^2} + \sqrt 3 \cos 4x = 4{\cos ^2}x – 1\\ \Leftrightarrow 2.{\left( {\dfrac{{\sqrt 2 }}{2}\cos 2x – \dfrac{{\sqrt 2 }}{2}\sin 2x} \right)^2} + \sqrt 3 .\cos 4x = 4{\cos ^2}x – 1\\ \Leftrightarrow 2.{\left( {\dfrac{{\sqrt 2 }}{2}} \right)^2}.{\left( {\cos 2x – \sin 2x} \right)^2} + \sqrt 3 .\cos 4x = 4{\cos ^2}x – 1\\ \Leftrightarrow 1.\left( {{{\cos }^2}2x – 2.\cos 2x.\sin 2x + {{\sin }^2}2x} \right) + \sqrt 3 \cos 4x = 4{\cos ^2}x – 1\\ \Leftrightarrow \left( {{{\sin }^2}2x + {{\cos }^2}2x} \right) – 2.\sin 2x.\cos 2x + \sqrt 3 \cos 4x = 4{\cos ^2}x – 1\\ \Leftrightarrow 1 – \sin 4x + \sqrt 3 .\cos 4x = 4{\cos ^2}x – 1\\ \Leftrightarrow – \sin 4x + \sqrt 3 \cos 4x = 2.\left( {2{{\cos }^2}x – 1} \right)\\ \Leftrightarrow – \dfrac{1}{2}\sin 4x + \dfrac{{\sqrt 3 }}{2}.\cos 4x = \cos 2x\\ \Leftrightarrow \cos 4x.\cos \dfrac{\pi }{6} – \sin 4x.\sin \dfrac{\pi }{6} = \cos 2x\\ \Leftrightarrow \cos \left( {4x + \dfrac{\pi }{6}} \right) = \cos 2x\\ \Leftrightarrow \left[ \begin{array}{l} 4x + \dfrac{\pi }{6} = 2x + k2\pi \\ 4x + \dfrac{\pi }{6} = – 2x + k2\pi \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l} x = – \dfrac{\pi }{{12}} + k\pi \\ x = – \dfrac{\pi }{{36}} + \dfrac{{k\pi }}{3} \end{array} \right.\\ 17,\\ \sqrt 3 \cos 5x – 2\sin 3x.\cos 2x – \sin x = 0\\ \Leftrightarrow \sqrt 3 \cos 5x – \left( {\sin \left( {3x + 2x} \right) + \sin \left( {3x – 2x} \right)} \right) – \sin x = 0\\ \Leftrightarrow \sqrt 3 \cos 5x – \sin 5x – \sin x – \sin x = 0\\ \Leftrightarrow \dfrac{{\sqrt 3 }}{2}.\cos 5x – \dfrac{1}{2}\sin 5x = \sin x\\ \Leftrightarrow \cos 5x.\cos \dfrac{\pi }{6} – \sin 5x.\sin \dfrac{\pi }{6} = \sin x\\ \Leftrightarrow \cos \left( {5x + \dfrac{\pi }{6}} \right) = \cos \left( {\dfrac{\pi }{2} – x} \right)\\ \Leftrightarrow \left[ \begin{array}{l} 5x + \dfrac{\pi }{6} = \dfrac{\pi }{2} – x + k2\pi \\ 5x + \dfrac{\pi }{6} = x – \dfrac{\pi }{2} + k2\pi \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l} x = \dfrac{\pi }{{18}} + \dfrac{{k\pi }}{3}\\ x = – \dfrac{\pi }{6} + \dfrac{{k\pi }}{2} \end{array} \right. \end{array}$$