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Câu hỏi: Giải phương trình ${2\cos x-\sqrt{2}=0}$.
A. ${x=\dfrac{\pi }{4}+k2\pi ,k\in \mathbb{Z}.}$
B. ${x=\pm \dfrac{\pi }{4}+k\pi ,k\in \mathbb{Z}.}$
C. ${x=\pm \dfrac{\pi }{4}+k2\pi ,k\in \mathbb{Z}.}$
D. ${x=-\dfrac{\pi }{4}+k2\pi ,k\in \mathbb{Z}.}$
A. ${x=\dfrac{\pi }{4}+k2\pi ,k\in \mathbb{Z}.}$
B. ${x=\pm \dfrac{\pi }{4}+k\pi ,k\in \mathbb{Z}.}$
C. ${x=\pm \dfrac{\pi }{4}+k2\pi ,k\in \mathbb{Z}.}$
D. ${x=-\dfrac{\pi }{4}+k2\pi ,k\in \mathbb{Z}.}$
$2\cos x-\sqrt{2}=0\Leftrightarrow \cos x=\dfrac{\sqrt{2}}{2}\Leftrightarrow \cos x=\cos \dfrac{\pi }{4}\Leftrightarrow x=\pm \dfrac{\pi }{4}+k2\pi ,k\in \mathbb{Z}$
Đáp án C.