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 …

0 bình luận về “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. $\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}$

   Bình luận