Tìm x∈(2π/5;6π/7) thỏa phương trình: cos7x−√3sin7x=−√2

Hướng dẫn giải:

Chia hai vế của (*) cho 2 ta được:

(*) \( \Leftrightarrow \frac{1}{2}\cos 7x-\frac{\sqrt{3}}{2}\sin 7x=-\frac{\sqrt{2}}{2}\Leftrightarrow -\sin \frac{\pi }{6}\cos 7x+\cos \frac{\pi }{6}\sin 7x=\frac{\sqrt{2}}{2} \)

 \( \Leftrightarrow \sin \left( 7x-\frac{\pi }{6} \right)=\sin \frac{\pi }{4}\Leftrightarrow \left[ \begin{align} & 7x-\frac{\pi }{6}=\frac{\pi }{4}+k2\pi  \\  & 7x-\frac{\pi }{6}=\pi -\frac{\pi }{4}+k2\pi  \\ \end{align} \right. \)

 \( \Leftrightarrow \left[ \begin{align}  & x=\frac{5\pi }{84}+\frac{k2\pi }{7} \\  & x=\frac{11\pi }{84}+\frac{h2\pi }{7} \\ \end{align} \right.,\text{ }k,h\in \mathbb{Z} \).

Do  \( x\in \left( \frac{2\pi }{5};\frac{6\pi }{7} \right) \) nên ta có:

+  \( \frac{2\pi }{5}<\frac{5\pi }{84}+\frac{k2\pi }{7}<\frac{6\pi }{7}\Leftrightarrow \frac{2}{5}<\frac{5}{84}+\frac{2k}{7}<\frac{6}{7}\xrightarrow{k\in \mathbb{Z}}k=2 \).

+ \(\frac{2\pi }{5}<\frac{11\pi }{84}+\frac{h2\pi }{7}<\frac{6\pi }{7}\Leftrightarrow \frac{2}{5}<\frac{11}{84}+\frac{2h}{7}<\frac{6}{7}\xrightarrow{h\in \mathbb{Z}}h\in \{1;2\}\).

Vậy:  \( x=\frac{5\pi }{84}+\frac{4\pi }{7}=\frac{53}{84}\pi \vee x=\frac{11\pi }{84}+\frac{2\pi }{7}=\frac{35\pi }{84}\vee x=\frac{11\pi }{84}+\frac{4\pi }{7}=\frac{59}{84}\pi \) .