Đáp án:
\(\begin{array}{l}
c,\\
x = \dfrac{\pi }{9} + \dfrac{1}{3}\arctan \dfrac{1}{2} + \dfrac{{k\pi }}{3}\,\,\,\,\left( {k \in Z} \right)\\
d,\\
x = - \dfrac{\pi }{{24}} + \dfrac{{k\pi }}{4}\,\,\,\,\left( {k \in Z} \right)
\end{array}\)
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
c,\\
DKXD:\,\,\cos \left( {3x - \dfrac{\pi }{3}} \right) \ne 0 \Leftrightarrow 3x - \dfrac{\pi }{3} \ne \dfrac{\pi }{2} + k\pi \\
\Leftrightarrow 3x \ne \dfrac{{5\pi }}{6} + k\pi \Leftrightarrow x \ne \dfrac{{5\pi }}{{18}} + \dfrac{{k\pi }}{3}\,\,\,\,\left( {k \in Z} \right)\\
\tan \left( {3x - \dfrac{\pi }{3}} \right) = \dfrac{1}{2}\\
\Leftrightarrow 3x - \dfrac{\pi }{3} = \arctan \dfrac{1}{2} + k\pi \\
\Leftrightarrow 3x = \dfrac{\pi }{3} + \arctan \dfrac{1}{2} + k\pi \\
\Leftrightarrow x = \dfrac{\pi }{9} + \dfrac{1}{3}\arctan \dfrac{1}{2} + \dfrac{{k\pi }}{3}\,\,\,\,\left( {k \in Z} \right)\\
d,\\
DKXD:\,\,\,\sin \left( {\dfrac{\pi }{3} - 4x} \right) \ne 0 \Leftrightarrow \dfrac{\pi }{3} - 4x \ne k\pi \\
\Leftrightarrow 4x \ne \dfrac{\pi }{3} + k\pi \Leftrightarrow x \ne \dfrac{\pi }{{12}} + \dfrac{{k\pi }}{4}\,\,\,\,\left( {k \in Z} \right)\\
\cot \left( {\dfrac{\pi }{3} - 4x} \right) = 0\\
\Leftrightarrow \cos \left( {\dfrac{\pi }{3} - 4x} \right) = 0\\
\Leftrightarrow \dfrac{\pi }{3} - 4x = \dfrac{\pi }{2} + k\pi \\
\Leftrightarrow 4x = - \dfrac{\pi }{6} + k\pi \\
\Leftrightarrow x = - \dfrac{\pi }{{24}} + \dfrac{{k\pi }}{4}\,\,\,\,\left( {k \in Z} \right)
\end{array}\)
