`~rai~`
\(\cos\left(3x-\dfrac{3\pi}{4}\right)=\dfrac{\sqrt{2}}{2}\\\Leftrightarrow \cos\left(3x-\dfrac{3\pi}{4}\right)=\cos\left(\dfrac{\pi}{4}\right)\\\Leftrightarrow \left[\begin{array}{I}3x-\dfrac{3\pi}{4}=\dfrac{\pi}{4}+k2\pi\\3x-\dfrac{3\pi}{4}+k2\pi\end{array}\right.\\\Leftrightarrow \left[\begin{array}{I}3x=\pi+k2\pi\\3x=\dfrac{\pi}{2}+k2\pi\end{array}\right.\\\Leftrightarrow \left[\begin{array}{I}x=\dfrac{\pi}{3}+k\dfrac{2\pi}{3}\\x=\dfrac{\pi}{6}+k\dfrac{2\pi}{3}.\end{array}\right.\quad(k\in\mathbb{Z})\)
