Giải phương trình: 1+cot2x=1−cos2x/sin^22x

Hướng dẫn giải:

Điều kiện:  \( \sin 2x\ne 0\Leftrightarrow \cos 2x\ne \pm 1 \).

Ta có: (*) \( \Leftrightarrow 1+\cot 2x=\frac{1-\cos 2x}{1-{{\cos }^{2}}2x}\Leftrightarrow 1+\cot 2x=\frac{1}{1+\cos 2x} \)

\(\Leftrightarrow \cot 2x=\frac{1}{1+\cos 2x}-1\Leftrightarrow \frac{\cos 2x}{\sin 2x}=\frac{-\cos 2x}{1+\cos 2x}\Leftrightarrow \cos 2x\left[ \frac{1}{\sin 2x}+\frac{1}{1+\cos 2x} \right]=0\)

 \( \Leftrightarrow \left[ \begin{align}  & \cos 2x=0\text{ }(\text{nhận }do\text{ }\cos 2x\ne \pm 1) \\  & \frac{1}{\sin 2x}=\frac{-1}{1+\cos 2x} \\ \end{align} \right. \) \( \Leftrightarrow \left[ \begin{align}  & \cos 2x=0 \\  & 1+\cos 2x=-\sin 2x \\ \end{align} \right. \)

 \( \Leftrightarrow \left[ \begin{align}  & \cos 2x=0 \\  & \sin 2x+\cos 2x=-1 \\ \end{align} \right. \) \( \Leftrightarrow \left[ \begin{align}  & \cos 2x=0 \\  & \sqrt{2}\sin \left( 2x+\frac{\pi }{4} \right)=-1 \\ \end{align} \right. \)

 \( \Leftrightarrow \left[ \begin{align}  & \cos 2x=0 \\  & \sin \left( 2x+\frac{\pi }{4} \right)=-\frac{\sqrt{2}}{2}=\sin \left( -\frac{\pi }{4} \right) \\ \end{align} \right. \) \( \Leftrightarrow \left[ \begin{align}  & 2x=\frac{\pi }{2}+k\pi  \\  & 2x+\frac{\pi }{4}=-\frac{\pi }{4}+k2\pi \vee 2x+\frac{\pi }{4}=\pi +\frac{\pi }{4}+k2\pi  \\ \end{align} \right. \)

 \( \Leftrightarrow \left[ \begin{align}  & x=\frac{\pi }{4}+\frac{k\pi }{2} \\  & x=-\frac{\pi }{4}+k\pi \vee x=\frac{\pi }{2}+k\pi \text{ }(\ell ) \\ \end{align} \right.\Leftrightarrow x=\frac{\pi }{4}+\frac{k\pi }{2} \).