Đáp án:
\[\left[ \begin{array}{l}
x = - \dfrac{\pi }{4} + k2\pi \\
x = \dfrac{{5\pi }}{4} + k2\pi
\end{array} \right.\,\,\,\,\left( {k \in Z} \right)\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
2\cos \left[ {\dfrac{\pi }{6}\left( {\sin x - 13 + \dfrac{{\sqrt 2 }}{2}} \right)} \right] = \sqrt 3 \\
\Leftrightarrow \cos \left[ {\dfrac{\pi }{6}\left( {\sin x - 13 + \dfrac{{\sqrt 2 }}{2}} \right)} \right] = \dfrac{{\sqrt 3 }}{2}\\
\Leftrightarrow \cos \left[ {\dfrac{\pi }{6}\left( {\sin x - 13 + \dfrac{{\sqrt 2 }}{2}} \right)} \right] = \cos \dfrac{\pi }{6}\\
\Leftrightarrow \left[ \begin{array}{l}
\dfrac{\pi }{6}\left( {\sin x - 13 + \dfrac{{\sqrt 2 }}{2}} \right) = \dfrac{\pi }{6} + k2\pi \\
\dfrac{\pi }{6}\left( {\sin x - 13 + \dfrac{{\sqrt 2 }}{2}} \right) = - \dfrac{\pi }{6} + k2\pi
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
\sin x - 13 + \dfrac{{\sqrt 2 }}{2} = 1 + 12k\\
\sin x - 13 + \dfrac{{\sqrt 2 }}{2} = - 1 + 12k
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
\sin x = 14 - \dfrac{{\sqrt 2 }}{2} + 12k\\
\sin x = 12 - \dfrac{{\sqrt 2 }}{2} + 12k
\end{array} \right.\,\,\,\,\,\left( {k \in Z} \right)\\
TH1:\,\,\,\sin x = 14 - \dfrac{{\sqrt 2 }}{2} + 12k\\
- 1 \le \sin x \le 1\\
\Leftrightarrow - 1 \le 14 - \dfrac{{\sqrt 2 }}{2} + 12k \le 1\\
\Leftrightarrow - 15 + \dfrac{{\sqrt 2 }}{2} \le 12k \le - 13 + \dfrac{{\sqrt 2 }}{2}\\
\Leftrightarrow - \dfrac{5}{4} + \dfrac{{\sqrt 2 }}{{24}} \le k \le - \dfrac{{13}}{{12}} + \dfrac{{\sqrt 2 }}{{24}}\\
k \in Z \Rightarrow Phương\,\,trình\,\,vô\,\,nghiệm\\
TH2:\,\,\,\sin x = 12 - \dfrac{{\sqrt 2 }}{2} + 12k\\
- 1 \le \sin x \le 1\\
\Leftrightarrow - 1 \le 12 - \dfrac{{\sqrt 2 }}{2} + 12k \le 1\\
\Leftrightarrow - 13 + \dfrac{{\sqrt 2 }}{2} \le 12k \le - 11 + \dfrac{{\sqrt 2 }}{2}\\
\Leftrightarrow - \dfrac{{13}}{{12}} + \dfrac{{\sqrt 2 }}{{24}} \le k \le - \dfrac{{11}}{{12}} + \dfrac{{\sqrt 2 }}{{24}}\\
k \in Z \Rightarrow k = - 1\\
\Rightarrow \sin x = - \dfrac{{\sqrt 2 }}{2}\\
\Leftrightarrow \sin x = \sin \dfrac{{ - \pi }}{4}\\
\Leftrightarrow \left[ \begin{array}{l}
x = - \dfrac{\pi }{4} + k2\pi \\
x = \pi - \left( { - \dfrac{\pi }{4}} \right) + k2\pi
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = - \dfrac{\pi }{4} + k2\pi \\
x = \dfrac{{5\pi }}{4} + k2\pi
\end{array} \right.\,\,\,\,\left( {k \in Z} \right)
\end{array}\)
