b: \(\Leftrightarrow\dfrac{1-\cos2x}{2}+\dfrac{1-\cos4x}{2}+\dfrac{1-\cos6x}{2}=\dfrac{3}{2}\)
\(\Leftrightarrow3-\cos2x-\cos4x-\cos6x=3\)
\(\Leftrightarrow\cos2x+\cos6x+\cos4x=0\)
\(\Leftrightarrow2\cdot\cos\left(\dfrac{6x+2x}{2}\right)\cdot\cos\left(\dfrac{6x-2x}{2}\right)+\cos2x=0\)
\(\Leftrightarrow\cos2x\left(2\cdot\cos4x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=\dfrac{\Pi}{2}+k\Pi\\\cos4x=-\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\Pi}{4}+\dfrac{k\Pi}{2}\\4x=-\dfrac{2}{3}\Pi+k2\Pi\\4x=\dfrac{2}{3}\Pi+k2\Pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\Pi}{4}+\dfrac{k\Pi}{2}\\x=-\dfrac{1}{6}\Pi+\dfrac{k\Pi}{2}\\x=\dfrac{1}{6}\Pi+\dfrac{k\Pi}{2}\end{matrix}\right.\)
