Đáp án:
$\begin{array}{l}
a)\tan 3x.\cot 5x = 1\\
\left( {dkxd:\left\{ \begin{array}{l}
\cos 3x \ne 0\\
\sin 5x \ne 0
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
x \ne \frac{\pi }{6} + \frac{{k\pi }}{3}\\
x \ne \frac{{k\pi }}{5}
\end{array} \right.} \right)\\
pt \Rightarrow \tan 3x = \frac{1}{{\cot 5x}}\\
\Rightarrow \tan 3x = \tan 5x\\
\Rightarrow 3x = 5x + k\pi \\
\Rightarrow x = \frac{{ - k\pi }}{2}\left( {ktmdk:x \ne \frac{\pi }{6} + \frac{{k\pi }}{3}} \right)\\
Vậy\,x \in \emptyset \\
b)2\sin \left( {x + \frac{\pi }{5}} \right) - \sqrt 3 = 0\\
\Rightarrow \sin \left( {x + \frac{\pi }{5}} \right) = \frac{{\sqrt 3 }}{2}\\
\Rightarrow \sin \left( {x + \frac{\pi }{5}} \right) = \sin \frac{\pi }{3}\\
\Rightarrow \left[ \begin{array}{l}
x + \frac{\pi }{5} = \frac{\pi }{3} + k2\pi \\
x + \frac{\pi }{5} = \pi - \frac{\pi }{3} + k2\pi
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x = \frac{{2\pi }}{{15}} + k2\pi \\
x = \frac{{7\pi }}{{15}} + k2\pi
\end{array} \right.\\
c)\cot \left( {x + \frac{\pi }{4}} \right) = 1\\
\Rightarrow x + \frac{\pi }{4} = \frac{\pi }{4} + k\pi \\
\Rightarrow x = k\pi
\end{array}$
