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

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

   $⇔sin(6x – \frac{\pi}{3}) = sin(\frac{\pi}{2} – x + \frac{\pi}{6})$

   $⇔sin(6x – \frac{\pi}{3}) = sin(\frac{2\pi}{3} – x)$

   $⇔ \left[ \begin{array}{l}6x – \frac{\pi}{3}=\frac{2\pi}{3} – x + k2\pi\\6x – \frac{\pi}{3}=\pi –  (\frac{2\pi}{3} – x) + k2\pi\end{array} \right. $

   $⇔\left[ \begin{array}{l}x=\frac{\pi}{7} + k\frac{2}{7}\pi\\x=\frac{2\pi}{15}+k\frac{2}{5}\pi\end{array} \right.$ ( $k ∈ Z )

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2. Giải thích các bước giải:

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

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

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

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

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

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