\(\Leftrightarrow\cos\left(x+\dfrac{\Pi}{4}\right)=\dfrac{1}{\sqrt{2}}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{\Pi}{4}=\dfrac{\Pi}{4}+k2\Pi\\x+\dfrac{\Pi}{4}=\dfrac{3}{4}\Pi+k2\Pi\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=k2\Pi\\x=\dfrac{1}{2}\Pi+k2\Pi\end{matrix}\right.\)
\(0\le x\le2\cdot\Pi\) nên \(x\in\left\{\dfrac{1}{2}\Pi;\Pi;\dfrac{3}{2}\Pi;2\Pi\right\}\)
=>CÓ 4 nghiệm