Đáp án:
\(\left[ \begin{array}{l}
x = \dfrac{{2\pi }}{3} + k2\pi \\
x = - \dfrac{{2\pi }}{3} + k2\pi
\end{array} \right.\,\,\,\,\left( {k \in Z} \right)\)
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\left( {2\cos x + 1} \right)\left( {\cos x - \sqrt 3 } \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
2\cos x + 1 = 0\\
\cos x - \sqrt 3 = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
\cos x = - \dfrac{1}{2}\\
\cos x = \sqrt 3
\end{array} \right.\\
- 1 \le \cos x \le 1 \Rightarrow \cos x = - \dfrac{1}{2}\\
\cos x = - \dfrac{1}{2}\\
\Leftrightarrow \cos x = \cos \dfrac{{2\pi }}{3}\\
\Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{{2\pi }}{3} + k2\pi \\
x = - \dfrac{{2\pi }}{3} + k2\pi
\end{array} \right.\,\,\,\,\left( {k \in Z} \right)
\end{array}\)
