2cos²x +2√3 sinx. cosx +1=3(sinx+√3cosx) 02/07/2021 Bởi Aaliyah 2cos²x +2√3 sinx. cosx +1=3(sinx+√3cosx)
$\begin{array}{l} 2{\cos ^2}x + 2\sqrt 3 \sin x\cos x + 1 = 3\left( {\sin x + \sqrt 3 \cos x} \right)\\ \Leftrightarrow 1 + \cos 2x + \sqrt 3 \sin 2x + 1 = 3\left( {\sin x + \sqrt 3 \cos x} \right)\\ \Leftrightarrow 2 = 3\left( {\sin x + \sqrt 3 \cos x} \right) – \left( {\cos 2x + \sqrt 3 \sin 2x} \right)\\ \Leftrightarrow 1 = 3\cos \left( {x – \dfrac{\pi }{6}} \right) – \cos \left( {2x – \dfrac{\pi }{3}} \right)\\ \Leftrightarrow 1 = 3\cos \left( {x – \dfrac{\pi }{6}} \right) – \left( {2{{\cos }^2}\left( {x – \dfrac{\pi }{6}} \right) – 1} \right)\\ \Leftrightarrow 3\cos \left( {x – \dfrac{\pi }{6}} \right) – 2{\cos ^2}\left( {x – \dfrac{\pi }{6}} \right) = 0\\ \Leftrightarrow \cos \left( {x – \dfrac{\pi }{6}} \right)\left( {3 – 2\cos \left( {x – \dfrac{\pi }{6}} \right)} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l} \cos \left( {x – \dfrac{\pi }{6}} \right) = 0\\ 3 – 2\cos \left( {x – \dfrac{\pi }{6}} \right) = 0 \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x – \dfrac{\pi }{6} = \dfrac{\pi }{2} + k\pi \\ \cos \left( {x – \dfrac{\pi }{6}} \right) = \dfrac{3}{2}(PTVN) \end{array} \right.\\ \Rightarrow x = \dfrac{{2\pi }}{3} + k\pi \left( {k \in \mathbb{Z}} \right)\\ \Rightarrow S = \left\{ {\dfrac{{2\pi }}{3} + k\pi |k \in \mathbb{Z}} \right\} \end{array}$ Bình luận
$\begin{array}{l} 2{\cos ^2}x + 2\sqrt 3 \sin x\cos x + 1 = 3\left( {\sin x + \sqrt 3 \cos x} \right)\\ \Leftrightarrow 1 + \cos 2x + \sqrt 3 \sin 2x + 1 = 3\left( {\sin x + \sqrt 3 \cos x} \right)\\ \Leftrightarrow 2 = 3\left( {\sin x + \sqrt 3 \cos x} \right) – \left( {\cos 2x + \sqrt 3 \sin 2x} \right)\\ \Leftrightarrow 1 = 3\cos \left( {x – \dfrac{\pi }{6}} \right) – \cos \left( {2x – \dfrac{\pi }{3}} \right)\\ \Leftrightarrow 1 = 3\cos \left( {x – \dfrac{\pi }{6}} \right) – \left( {2{{\cos }^2}\left( {x – \dfrac{\pi }{6}} \right) – 1} \right)\\ \Leftrightarrow 3\cos \left( {x – \dfrac{\pi }{6}} \right) – 2{\cos ^2}\left( {x – \dfrac{\pi }{6}} \right) = 0\\ \Leftrightarrow \cos \left( {x – \dfrac{\pi }{6}} \right)\left( {3 – 2\cos \left( {x – \dfrac{\pi }{6}} \right)} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l} \cos \left( {x – \dfrac{\pi }{6}} \right) = 0\\ 3 – 2\cos \left( {x – \dfrac{\pi }{6}} \right) = 0 \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x – \dfrac{\pi }{6} = \dfrac{\pi }{2} + k\pi \\ \cos \left( {x – \dfrac{\pi }{6}} \right) = \dfrac{3}{2}(PTVN) \end{array} \right.\\ \Rightarrow x = \dfrac{{2\pi }}{3} + k\pi \left( {k \in \mathbb{Z}} \right)\\ \Rightarrow S = \left\{ {\dfrac{{2\pi }}{3} + k\pi |k \in \mathbb{Z}} \right\} \end{array}$