Giải pt a. Sin4x.cot3x=0 b. (Cot x/3 -1)(tan x/2 +1)=0 c. tan(x-30°)cos(2x-150°)=0

0 bình luận về “Giải pt a. Sin4x.cot3x=0 b. (Cot x/3 -1)(tan x/2 +1)=0 c. tan(x-30°)cos(2x-150°)=0”

1. Giải thích các bước giải:

   Ta có:

   \(\begin{array}{l}
   a,\\
   DK:\,\,\,sin3x \ne 0 \Leftrightarrow 3x \ne k\pi  \Leftrightarrow x \ne \dfrac{{k\pi }}{3}\\
   \sin 4x.cot3x = 0\\
    \Leftrightarrow \sin 4x.\dfrac{{\cos 3x}}{{\sin 3x}} = 0\\
    \Leftrightarrow \left[ \begin{array}{l}
   \sin 4x = 0\\
   \cos 3x = 0
   \end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
   4x = k\pi \\
   3x = \dfrac{\pi }{2} + k\pi 
   \end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
   x = \dfrac{{k\pi }}{4}\\
   x = \dfrac{\pi }{6} + \dfrac{{k\pi }}{3}
   \end{array} \right.\\
   b,\\
   DK:\,\,\,\,\left\{ \begin{array}{l}
   \sin \dfrac{x}{3} \ne 0\\
   \cos \dfrac{x}{2} \ne 0
   \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
   \dfrac{x}{3} \ne k\pi \\
   \dfrac{x}{2} \ne \dfrac{\pi }{2} + k\pi 
   \end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
   x \ne k3\pi \\
   x \ne \pi  + k2\pi 
   \end{array} \right.\\
   \left( {\cot \dfrac{x}{3} – 1} \right).\left( {\tan \dfrac{x}{2} + 1} \right) = 0\\
    \Leftrightarrow \left[ \begin{array}{l}
   \cot \dfrac{x}{3} = 1\\
   \tan \dfrac{x}{2} =  – 1
   \end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
   \dfrac{x}{3} = \dfrac{\pi }{4} + k\pi \\
   \dfrac{x}{2} =  – \dfrac{\pi }{4} + k\pi 
   \end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
   x = \dfrac{{3\pi }}{4} + k3\pi \\
   x =  – \dfrac{\pi }{2} + k2\pi 
   \end{array} \right.\\
   c,\\
   DK:\,\,\,\cos \left( {x – 30^\circ } \right) \ne 0 \Leftrightarrow x – 30^\circ  \ne 90^\circ  + k.180^\circ  \Leftrightarrow x \ne 120^\circ  + k.180^\circ \\
   \tan \left( {x – 30^\circ } \right).\cos \left( {2x – 150^\circ } \right) = 0\\
    \Leftrightarrow \left[ \begin{array}{l}
   \tan \left( {x – 30^\circ } \right) = 0\\
   \cos \left( {2x – 150^\circ } \right) = 0
   \end{array} \right.\\
    \Leftrightarrow \left[ \begin{array}{l}
   x – 30^\circ  = k.180^\circ \\
   2x – 150^\circ  = 90^\circ  + k.180^\circ 
   \end{array} \right.\\
    \Leftrightarrow \left[ \begin{array}{l}
   x = 30^\circ  + k.180^\circ \\
   x = 120^\circ  + k.90^\circ 
   \end{array} \right.\\
    \Rightarrow \left[ \begin{array}{l}
   x = 30^\circ  + k.180^\circ \\
   \left\{ \begin{array}{l}
   x = 120^\circ  + k.90^\circ \\
   x \ne 120^\circ  + k.180^\circ 
   \end{array} \right.
   \end{array} \right.
   \end{array}\)

   Bình luận