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Câu hỏi: Nghiệm của phương trình ${\cos \left( x+\dfrac{\pi }{4} \right)=\dfrac{\sqrt{2}}{2}}$ là
A. ${\left[x=k2px=-p2+kpx=k2px=-p2+kp\right.\left( k\in \mathbb{Z} \right)}$.[/OPTION_A]
B. ${\left[ x=kpx=-p2+kpx=kpx=-p2+kp \right.\left( k\in \mathbb{Z} \right)}$.
C. ${\left[ x=kpx=-p2+k2px=kpx=-p2+k2p \right.\left( k\in \mathbb{Z} \right)}$.
D. ${\left[ x=k2px=-p2+k2px=k2px=-p2+k2p \right.\left( k\in \mathbb{Z} \right)}$.
A. ${\left[x=k2px=-p2+kpx=k2px=-p2+kp\right.\left( k\in \mathbb{Z} \right)}$.[/OPTION_A]
B. ${\left[ x=kpx=-p2+kpx=kpx=-p2+kp \right.\left( k\in \mathbb{Z} \right)}$.
C. ${\left[ x=kpx=-p2+k2px=kpx=-p2+k2p \right.\left( k\in \mathbb{Z} \right)}$.
D. ${\left[ x=k2px=-p2+k2px=k2px=-p2+k2p \right.\left( k\in \mathbb{Z} \right)}$.
Phương trình $\cos \left( x+\dfrac{\pi }{4} \right)=\dfrac{\sqrt{2}}{2}\Leftrightarrow \cos \left( x+\dfrac{\pi }{4} \right)=\cos \left( \dfrac{\pi }{4} \right)\Rightarrow \left[ \begin{aligned}
& x=k2\pi \\
& x=-\dfrac{\pi }{2}+k2\pi \\
\end{aligned} \right.\left( k\in \mathbb{Z} \right)$
& x=k2\pi \\
& x=-\dfrac{\pi }{2}+k2\pi \\
\end{aligned} \right.\left( k\in \mathbb{Z} \right)$
Đáp án D.