Trên khoảng \(\left( {\frac{\pi }{2};2\pi } \right)\), phương trình \(\cos \left( {\frac{\pi }{6} - 2\pi } \right) = \sin x\) có bao nhiêu nghiệm ?

\(\cos \left( {\frac{\pi }{6} - 2\pi } \right) = \sin x\)

\(\begin{array}{l} \Leftrightarrow \cos \left( {\frac{\pi }{6} - 2x} \right) = \cos \left( {\frac{\pi }{2} - x} \right)\\ \Leftrightarrow \frac{\pi }{6} - 2x = \pm \left( {\frac{\pi }{2} - x} \right) + k2\pi \,\,\,\,\left( {k \in \mathbb{Z}} \right)\end{array}\)

\( \Leftrightarrow \left[ \begin{array}{l}x = - \frac{\pi }{3} + k2\pi \\3x = \frac{{2\pi }}{3} + k\pi \end{array} \right.\,\,\left( {k \in \mathbb{Z}} \right)\)

\( \Leftrightarrow \left[ \begin{array}{l}x = - \frac{\pi }{3} + k2\pi \,\,\,\,\,\left( 1 \right)\\x = \frac{{2\pi }}{9} + 2k\frac{\pi }{3}\,\,\,\,\left( 2 \right)\end{array} \right.\,\,\left( {k \in \mathbb{Z}} \right)\)

Do \(x \in \left( {\frac{\pi }{2};2\pi } \right)\)

Nên (1) \( \Leftrightarrow \frac{1}{2} < - \frac{1}{3} + 2k < 2 \Leftrightarrow \frac{5}{{12}} < k < \frac{7}{6} \Rightarrow k = \left\{ 1 \right\}\), do k nguyên.

(2) \[ \Leftrightarrow \frac{1}{2} < \frac{2}{9} + \frac{{2k}}{3} < 2 \Leftrightarrow \frac{1}{2} < \frac{2}{3} + 2k < 6\]

\( \Leftrightarrow \frac{1}{6} < k < \frac{8}{3} \Rightarrow k \in \left\{ {1;2} \right\}\), do k nguyên.

Vậy có 3 nghiệm thỏa mãn yêu cầu.