2. sin(6x - $\frac{\pi}{3}$ ) = cos(x - $\frac{\pi}{6}$ )

Giải thích các bước giải:

$\sin{\left ( 6x - \dfrac{\pi}{3} \right )} = \cos{\left ( x - \dfrac{\pi}{6} \right )}$

$\Leftrightarrow \cos{\left ( \dfrac{\pi}{2} - 6x + \dfrac{\pi}{3} \right )} = \cos{\left ( x - \dfrac{\pi}{6} \right )}$

$\Leftrightarrow \cos{\left ( \dfrac{5\pi}{6} - 6x \right )} = \cos{\left ( x - \dfrac{\pi}{6} \right )}$

$\Leftrightarrow \left[ \begin{array}{l}\dfrac{5\pi}{6} - 6x = x - \dfrac{\pi}{6} + k2\pi\\\dfrac{5\pi}{6} - 6x = \dfrac{\pi}{6} - x + k2\pi\end{array} \right.$

$\Leftrightarrow \left[ \begin{array}{l}x = -\dfrac{\pi}{7} - \dfrac{k2\pi}{7}\\x = \dfrac{2\pi}{15} - \dfrac{k2\pi}{5}\end{array} \right.$