|
|
Điều kiện $\cot x \ne 1,\sin 2x \ne 0.$
$\Leftrightarrow \dfrac{1}{\dfrac{\sin x}{\cos x}+\dfrac{\cos 2x}{\sin 2x}}=\dfrac{\sqrt{2}(\cos x-\sin x)}{\dfrac{\cos x}{\sin x}-1}$
$\Leftrightarrow \dfrac{1}{\dfrac{2\sin^2 x+\cos 2x}{\sin 2x}}=\dfrac{\sqrt{2}(\cos x-\sin x)}{\dfrac{\cos x-\sin x}{\sin x}}$
$\Leftrightarrow \dfrac{\sin 2x}{2\sin^2 x+\cos 2x}=\sqrt{2}\sin x$
$\Leftrightarrow \dfrac{\sin 2x}{1-\cos 2x+\cos 2x}=\sqrt{2}\sin x$
$\Leftrightarrow \sin 2x=\sqrt{2}\sin x$
$\Leftrightarrow \cos x=1/\sqrt{2}$, vì $\sin x \ne 0.$
|