Nghiệm của phương trình \(\cos\left(x+\dfrac{\pi}{4}\right)=\dfrac{\sqrt{2}}{2}\) là
| \(\left[\begin{array}{l}x=k2\pi\\ x=-\dfrac{\pi}{2}+k\pi\end{array}\right.\,(k\in\mathbb{Z})\) | |
| \(\left[\begin{array}{l}x=k\pi\\ x=-\dfrac{\pi}{2}+k\pi\end{array}\right.\,(k\in\mathbb{Z})\) | |
| \(\left[\begin{array}{l}x=k\pi\\ x=-\dfrac{\pi}{2}+k2\pi\end{array}\right.\,(k\in\mathbb{Z})\) | |
| \(\left[\begin{array}{l}x=k2\pi\\ x=-\dfrac{\pi}{2}+k2\pi\end{array}\right.\,(k\in\mathbb{Z})\) |
Chọn phương án D.
\(\begin{aligned}
\cos\left(x+\dfrac{\pi}{4}\right)=\dfrac{\sqrt{2}}{2}\Leftrightarrow&\left[\begin{array}{l}x+\dfrac{\pi}{4}=\dfrac{\pi}{4}+k2\pi\\ x+\dfrac{\pi}{4}=-\dfrac{\pi}{4}+k2\pi\end{array}\right.\\
\Leftrightarrow&\left[\begin{array}{l}x=k2\pi\\ x=-\dfrac{\pi}{2}+k2\pi\end{array}\right.\,(k\in\mathbb{Z}).
\end{aligned}\)