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

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}\)