2sin(3x-15°)=căn3 Cos 2x +9cossx-10=0 Sin2x-căn3 cos2x=1 1+sinx-cos2x -sin3x=0 Giúp e vs e đag kiểm tra ...
1 câu trả lời
$\begin{array}{l} 1)pt \Leftrightarrow \sin \left( {3x - {{15}^0}} \right) = \frac{{\sqrt 3 }}{2}\\ \Leftrightarrow \left[ \begin{array}{l} 3x - {15^0} = {60^0} + k{360^0}\\ 3x - {15^0} = {120^0} + k{360^0} \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l} 3x = {75^0} + k{360^0}\\ 3x = {135^0} + k{360^0} \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l} x = {15^0} + k{120^0}\\ x = {45^0} + k{120^0} \end{array} \right.\\ 2)\cos 2x + 9\cos x - 10 = 0\\ \Leftrightarrow 2{\cos ^2}x - 1 + 9\cos x - 10 = 0\\ \Leftrightarrow 2{\cos ^2}x + 9\cos x - 11 = 0\\ \Leftrightarrow \left[ \begin{array}{l} \cos x = 1\\ \cos x = - \frac{{11}}{2}\left( {VN} \right) \end{array} \right. \Leftrightarrow x = k2\pi \\ 3)\sin 2x - \sqrt 3 \cos 2x = 1\\ \Leftrightarrow \frac{1}{2}\sin 2x - \frac{{\sqrt 3 }}{2}\cos 2x = \frac{1}{2}\\ \Leftrightarrow \sin \left( {2x - \frac{\pi }{3}} \right) = \sin \frac{\pi }{6}\\ \Leftrightarrow \left[ \begin{array}{l} 2x - \frac{\pi }{3} = \frac{\pi }{6} + k2\pi \\ 2x - \frac{\pi }{3} = \frac{{5\pi }}{6} + k2\pi \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l} 2x = \frac{\pi }{2} + k2\pi \\ 2x = \frac{{7\pi }}{6} + k2\pi \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l} x = \frac{\pi }{4} + k\pi \\ x = \frac{{7\pi }}{{12}} + k\pi \end{array} \right.\\ 4)1 + \sin x - \cos 2x - \sin 3x = 0\\ \Leftrightarrow 1 + \sin x - \left( {1 - 2{{\sin }^2}x} \right) - \left( {3\sin x - 4{{\sin }^3}x} \right) = 0\\ \Leftrightarrow 1 + \sin x - 1 + 2{\sin ^2}x - 3\sin x + 4{\sin ^3}x = 0\\ \Leftrightarrow 4{\sin ^3}x + 2{\sin ^2}x - 2\sin x = 0\\ \Leftrightarrow \sin x\left( {2{{\sin }^2}x + \sin x - 1} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l} \sin x = 0\\ \sin x = - 1\\ \sin x = \frac{1}{2} \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x = k\pi \\ x = - \frac{\pi }{2} + k\pi \\ x = \frac{\pi }{6} + k2\pi \\ x = \frac{{5\pi }}{6} + k2\pi \end{array} \right. \end{array}$