Đáp án:
8) \(x \ne \dfrac{\pi }{{12}} + \dfrac{{k\pi }}{2}\)
Giải thích các bước giải:
\(\begin{array}{l}
1)DK:\sin x \ne 0\\
\to x \ne k\pi \left( {k \in Z} \right)\\
2)DK:{\sin ^2}x \ne 1\\
\to \left\{ \begin{array}{l}
\sin x \ne 1\\
\sin x \ne - 1
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x \ne \dfrac{\pi }{2} + k2\pi \\
x \ne - \dfrac{\pi }{2} + k2\pi
\end{array} \right.\\
\to x \ne \dfrac{\pi }{2} + k2\pi \left( {k \in Z} \right)\\
3)DK:\cos \left( {2x - \dfrac{\pi }{3}} \right) \ne 0\\
\to 2x - \dfrac{\pi }{3} \ne \dfrac{\pi }{2} + k\pi \\
\to x \ne \dfrac{{5\pi }}{{12}} + \dfrac{{k\pi }}{2}\left( {k \in Z} \right)\\
4)DK:\sin \left( {x + \dfrac{\pi }{3}} \right) \ne 0\\
\to x + \dfrac{\pi }{3} \ne k\pi \\
\to x \ne - \dfrac{\pi }{3} + k\pi \left( {k \in Z} \right)\\
5)DK:\left\{ \begin{array}{l}
\cos x \ge - 1\left( {ld} \right)\\
\cos x \ne 0
\end{array} \right.\\
\to x \ne \dfrac{\pi }{2} + k\pi \left( {k \in Z} \right)\\
6)DK:\left\{ \begin{array}{l}
1 \ge \sin x\\
\sin x \ge - 1
\end{array} \right.\left( {ld} \right)\\
\to DK:\forall x\\
7)DK:2 \ge \sin \dfrac{x}{2}\left( {ld} \right)\\
\to DK:\forall x\\
8)DK:\cos \left( {2x + \dfrac{\pi }{3}} \right) \ne 0\\
\to 2x + \dfrac{\pi }{3} \ne \dfrac{\pi }{2} + k\pi \\
\to x \ne \dfrac{\pi }{{12}} + \dfrac{{k\pi }}{2}\left( {k \in Z} \right)
\end{array}\)
