Phương trình \(\cos x+\sin x=\frac{\cos 2 x}{1-\sin 2 x}\) có nghiệm là:
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Lời giải:
Báo saiĐK: \(\sin 2 x \neq 1\)
\(\begin{array}{l} \cos x+\sin x=\frac{\cos 2 x}{1-\sin 2 x} \Leftrightarrow \cos x+\sin x=\frac{\cos ^{2} x-\sin ^{2} x}{(\sin x-\cos x)^{2}} \\ \Leftrightarrow \cos x+\sin x=\frac{(\cos x-\sin x)(\cos x+\sin x)}{(\sin x-\cos x)^{2}} \\ \Leftrightarrow \cos x+\sin x=-\frac{\cos x+\sin x}{\sin x-\cos x} \Leftrightarrow(\cos x+\sin x)\left(1+\frac{1}{\sin x-\cos x}\right)=0 \end{array}\)
\(\Leftrightarrow\left[\begin{array}{l} \cos x+\sin x=0 \\ \sin x-\cos x=-1 \end{array} \Leftrightarrow\left[\begin{array}{l} \sqrt{2} \sin \left(x+\frac{\pi}{4}\right)=0 \\ \sqrt{2} \sin \left(x-\frac{\pi}{4}\right)=-1 \end{array}\right.\right.\)
\(\Leftrightarrow\left[\begin{array}{l} x+\frac{\pi}{4}=k \pi \\ x-\frac{\pi}{4}=-\frac{\pi}{4}+k 2 \pi(k \in \mathbb{Z}) \\ x-\frac{\pi}{4}=\frac{5 \pi}{4}+k 2 \pi \end{array}\right.\)
\(\Leftrightarrow\left[\begin{array}{l} x=-\frac{\pi}{4}+k \pi \\ x=k 2 \pi \\ x=\frac{3 \pi}{2}+k 2 \pi \end{array}(k \in \mathbb{Z})\right.\)
\(\Leftrightarrow\left[\begin{array}{l} x=\frac{3 \pi}{4}+k \pi \\ x=-\frac{\pi}{2}+k 2 \pi(k \in \mathbb{Z}) \\ x=k 2 \pi \end{array}\right.\)