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Câu hỏi: Số nghiệm của phương trình ${{\cos }^{2}}x-\sin 2x=\sqrt{2}+{{\cos }^{2}}\left( \dfrac{\pi }{2}+x \right)$ trên khoảng $\left( 0;4\pi \right)$ là
A. 3.
B. 5.
C. 4.
D. 2.
A. 3.
B. 5.
C. 4.
D. 2.
Ta có ${{\cos }^{2}}x-\sin 2x=\sqrt{2}+{{\cos }^{2}}\left( \dfrac{\pi }{2}+x \right)\Leftrightarrow {{\cos }^{2}}x-\sin 2x=\sqrt{2}+{{\sin }^{2}}x$
$\Leftrightarrow \cos 2x-\sin 2x=\sqrt{2}\Leftrightarrow \sin 2x-\cos 2x=-\sqrt{2}\Leftrightarrow \sin \left( 2x-\dfrac{\pi }{4} \right)=-1$
$\Leftrightarrow 2x-\dfrac{\pi }{4}=-\dfrac{\pi }{2}+k2\pi \Leftrightarrow x=-\dfrac{\pi }{8}+k\pi $ $\left( k\in \mathbb{Z} \right)$
Mà $x\in \left( 0;4\pi \right)\Leftrightarrow 0<-\dfrac{\pi }{8}+k\pi <4\pi \Leftrightarrow \dfrac{1}{8}<k<\dfrac{33}{8}\xrightarrow{k\in \mathbb{Z}}k=\left\{ 1;2;3;4 \right\}$.
$\Leftrightarrow \cos 2x-\sin 2x=\sqrt{2}\Leftrightarrow \sin 2x-\cos 2x=-\sqrt{2}\Leftrightarrow \sin \left( 2x-\dfrac{\pi }{4} \right)=-1$
$\Leftrightarrow 2x-\dfrac{\pi }{4}=-\dfrac{\pi }{2}+k2\pi \Leftrightarrow x=-\dfrac{\pi }{8}+k\pi $ $\left( k\in \mathbb{Z} \right)$
Mà $x\in \left( 0;4\pi \right)\Leftrightarrow 0<-\dfrac{\pi }{8}+k\pi <4\pi \Leftrightarrow \dfrac{1}{8}<k<\dfrac{33}{8}\xrightarrow{k\in \mathbb{Z}}k=\left\{ 1;2;3;4 \right\}$.
Đáp án C.