Đáp án: \(\left[ \begin{array}{l}x = \frac{π}{6} + kπ\\x = -\frac{π}{12} + k\frac{π}{2}\end{array} \right.\) `(k ∈ ZZ)`
Giải thích các bước giải:
`ĐKXĐ: x ne k(π)/2`
`8sin x = (\sqrt{3})/(cos x) + 1/(sin x)`
`<=> 4sin x(2.cos x. sin x) = sqrt{3}.sin x + cos x`
`<=> 4sin x.sin 2x = sqrt{3}.sin x + cos x`
`<=> -2(cos 3x - cos x) = sqrt{3}sin x + cos x`
`<=> 2cos 3x = sqrt{3}sin x + cos x`
`<=> cos 3x = (\sqrt{3})/(2)sin x + 1/(2)cos x`
`<=> cos 3x = cos (x + π/3)`
`<=>` \(\left[ \begin{array}{l}3x = x + \frac{π}{3} + k2π\\3x = -x - \frac{π}{3} + k2π\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}x = \frac{π}{6} + kπ\\x = -\frac{π}{12} + k\frac{π}{2}\end{array} \right.\) `(k ∈ ZZ)`
