giải pt lượng giác a) (Cotx +1)× sin3x =0 b) sinx ( x-2pi/3)+ cos2x=0

Giải thích các bước giải:

 Ta có:

\(\begin{array}{l}
a,\\
DK:\,\,\,\,\sin x \ne 0 \Leftrightarrow x \ne k\pi \\
\left( {\cot x + 1} \right).\sin 3x = 0\\
 \Leftrightarrow \left( {\dfrac{{\cos x}}{{\sin x}} + 1} \right).\sin 3x = 0\\
 \Leftrightarrow \dfrac{{\cos x + \sin x}}{{\sin x}}.\sin 3x = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
\cos x + \sin x = 0\\
\sin 3x = 0
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
\sqrt 2 .\sin \left( {x + \dfrac{\pi }{4}} \right) = 0\\
3x = k\pi 
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
\sin \left( {x + \dfrac{\pi }{4}} \right) = 0\\
x = \dfrac{{k\pi }}{3}
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x + \dfrac{\pi }{4} = k\pi \\
\left\{ \begin{array}{l}
x = \dfrac{{k\pi }}{3}\\
x \ne k\pi 
\end{array} \right.
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x =  – \dfrac{\pi }{4} + k\pi \\
x = \dfrac{{k\pi }}{3}\,\,\,\,\,\left( {k \ne 3l} \right)
\end{array} \right.\,\,\,\,\,\,\,\,\left( {k,l \in Z} \right)\\
b,\\
\sin \left( {x – \dfrac{{2\pi }}{3}} \right) + \cos 2x = 0\\
 \Leftrightarrow \cos 2x =  – \sin \left( {x – \dfrac{{2\pi }}{3}} \right)\\
 \Leftrightarrow \cos 2x = \sin \left( {\dfrac{{2\pi }}{3} – x} \right)\\
 \Leftrightarrow \cos 2x = \cos \left[ {\dfrac{\pi }{2} – \left( {\dfrac{{2\pi }}{3} – x} \right)} \right]\\
 \Leftrightarrow \cos 2x = \cos \left( {x – \dfrac{\pi }{6}} \right)\\
 \Leftrightarrow \left[ \begin{array}{l}
2x = x – \dfrac{\pi }{6} + k2\pi \\
2x = \dfrac{\pi }{6} – x + k2\pi 
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x =  – \dfrac{\pi }{6} + k2\pi \\
x = \dfrac{\pi }{{18}} + \dfrac{{k2\pi }}{3}
\end{array} \right.\,\,\,\left( {k \in Z} \right)
\end{array}\)