b: \(\Leftrightarrow\sqrt{3}\cdot\cos6x-2\cdot\dfrac{1}{2}\cdot\left(\sin6x+\sin2x\right)-\sin2x=0\)
\(\Leftrightarrow\sqrt{3}\cdot\cos6x-\sin6x-\sin2x-\sin2x=0\)
\(\Leftrightarrow\cos6x\cdot\dfrac{\sqrt{3}}{2}-\sin6x\cdot\dfrac{1}{2}=\sin2x\)
\(\Leftrightarrow\sin6x\cdot\dfrac{1}{2}-\cos6x\cdot\dfrac{\sqrt{3}}{2}=\sin\left(-2x\right)\)
\(\Leftrightarrow\sin\left(6x-\dfrac{\Pi}{3}\right)=\sin\left(-2x\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}6x-\dfrac{\Pi}{3}=-2x+k2\Pi\\6x-\dfrac{\Pi}{3}=\Pi+2x+k2\Pi\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{8}\left(k2\Pi+\dfrac{\Pi}{3}\right)\\4x=\dfrac{4}{3}\Pi+k2\Pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{k\Pi}{4}+\dfrac{\Pi}{24}\\x=\dfrac{1}{3}\Pi+\dfrac{k\Pi}{2}\end{matrix}\right.\)
