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Ta tìm nghiệm của (*)
$ \begin{array}{l}
(*) \Leftrightarrow \frac{{\sqrt 3 }}{2}\sin 7x - \frac{1}{2}c{\rm{os}}7x = \frac{{\sqrt 2 }}{2}\\
\,\,\,\,\,\,\, \Leftrightarrow \sin \left( {7x - \frac{\pi }{6}} \right) = \sin \frac{\pi }{4}\\
\,\,\,\,\,\,\, \Leftrightarrow \left[ \begin{array}{l}
x = \frac{{5\pi }}{{84}} + \frac{{k2\pi }}{7}\\
x = \frac{{11\pi }}{{84}} + \frac{{k2\pi }}{7}
\end{array} \right.;(k \in Z)
\end{array} $
Tìm k là số nguyên sao cho nghiệm thỏa mãn giả thiết:
Ta xét 2 trường hợp:
$ \begin{array}{l}
{\rm{TH}}1:\,\,\,x = \frac{{5\pi }}{{84}} + \frac{{k2\pi }}{7}\\
\Rightarrow \frac{{2\pi }}{5} < \frac{{5\pi }}{{84}} + \frac{{k2\pi }}{7} < \frac{{6\pi }}{7} \Leftrightarrow \frac{2}{5} - \frac{5}{{84}} < \frac{{2k}}{7} < \frac{6}{7} - \frac{5}{{84}}\\
\Leftrightarrow k = 2 \Leftrightarrow {x_1} = \frac{{53\pi }}{{84}}
\end{array} $
$ \begin{array}{l}
{\rm{TH2}}:\,\,\,x = \frac{{11\pi }}{{84}} + \frac{{k2\pi }}{7}\\
\Rightarrow \frac{{2\pi }}{5} < \frac{{11\pi }}{{84}} + \frac{{k2\pi }}{7} < \frac{{6\pi }}{7} \Leftrightarrow \frac{2}{5} - \frac{{11}}{{84}} < \frac{{2k}}{7} < \frac{6}{7} - \frac{{11}}{{84}}\\
\Leftrightarrow k = 1,2 \Leftrightarrow {x_2} = \frac{{35\pi }}{{84}};{x_3} = \frac{{59\pi }}{{84}}
\end{array} $
Vậy nghiệm cần tìm là: $ {x_1} = \frac{{53\pi }}{{84}};\,\,{x_2} = \frac{{35\pi }}{{84}};\,\,{x_3} = \frac{{59\pi }}{{84}} $
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