\(\frac{2\cos2x}{1-\sin2x}=0\Leftrightarrow\hept{\begin{cases}\cos2x=0\\1-\sin2x\ne0\end{cases}}\)
\(\cos2x=0\Leftrightarrow2x\pm\frac{\pi}{2}+k2\pi\)
\(\Leftrightarrow x=\pm\frac{\pi}{4}+k\pi\)
Với \(x=\frac{\pi}{4}+k\pi\Rightarrow2x=\frac{\pi}{2}+k2\pi\)
\(\sin2x=\sin\left(\frac{\pi}{2}+k2\pi\right)=1\) vi phạm điều kiện \(1-\sin2x\ne0\)
Do đó ta loại nghiệm \(x=\frac{\pi}{4}+k\pi\) của phương trình cos2x = 0
Vậy \(\frac{2\cos2x}{1-\sin2x}=0\Leftrightarrow x=-\frac{\pi}{4}+k\pi,k\in Z\)
Thanks Kikyo =) Nhưng t ko tíc được sory
\(\frac{2\cos2x}{1-\sin2x}=0\)
\(\Leftrightarrow\frac{2\left(\cos^2x-\sin^2x\right)}{\left(\cos x-\sin x\right)^2}=0\)
\(\Leftrightarrow\frac{2\left(\cos x-\sin x\right)\left(\cos x+\sin x\right)}{\left(\cos x-\sin x\right)\left(\cos x-\sin x\right)}=0\)
\(\Leftrightarrow\frac{2\left(\cos x+\sin x\right)}{\cos x-\sin x}=0\rightarrow\text{ĐK: }x\ne\frac{\pi}{4}+k\pi\)
\(\Leftrightarrow\cos x+\sin x=0\)
\(\Leftrightarrow\sqrt{z}\sin\left(x+\frac{\pi}{a}\right)=0\)
\(\Leftrightarrow x+\frac{\pi}{4}=k\pi\)
\(\Leftrightarrow x=\frac{-\pi}{4}+k\pi\)