Đáp án:
\(x \in \left\{ {{{2\pi } \over 3};{\pi \over 7};{{3\pi } \over 7};{{5\pi } \over 7}} \right\}\)
Giải thích các bước giải:
\(\eqalign{
& {{\sin 5x} \over {\sin x}} = 2\cos x\,\,\left( {x \ne k\pi } \right) \cr
& \Leftrightarrow \sin 5x = 2\sin x\cos x \cr
& \Leftrightarrow \sin 5x = \sin 2x \cr
& \Leftrightarrow \left[ \matrix{
5x = 2x + k2\pi \hfill \cr
5x = \pi - 2x + k2\pi \hfill \cr} \right. \cr
& \Leftrightarrow \left[ \matrix{
3x = k2\pi \hfill \cr
7x = \pi + k2\pi \hfill \cr} \right. \cr
& \Leftrightarrow \left[ \matrix{
x = {{k2\pi } \over 3} \hfill \cr
x = {\pi \over 7} + {{k2\pi } \over 7} \hfill \cr} \right. \cr
& x \in \left( {0;\pi } \right) \Rightarrow x \in \left\{ {{{2\pi } \over 3};{\pi \over 7};{{3\pi } \over 7};{{5\pi } \over 7}} \right\} \cr} \)
