Giải phương trình: \(\sqrt{2}\sin 2x=\frac{\sin x-\cos 3x+2\cos 2x\cos x}

Giải phương trình \(\sqrt{2}\sin 2x=\frac{\sin x-\cos 3x+2\cos 2x\cos x}{\tan \left ( \frac{\pi }{4}-x \right )\tan \left ( \frac{\pi }{4}+x \right )}.\; \; (1)\)

Điều kiện:

\(\left\{\begin{matrix} \tan \left ( \frac{\pi }{4}-x \right )\tan \left ( \frac{\pi }{4}+x \right )\neq 0\\ \cos \left ( \frac{\pi }{4}-x \right )\cos \left ( \frac{\pi}{4}+x \right )\neq 0 \end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} \! \! \! \! \! \! \! \! \! \! \! \! \! \! \! \! \! \! \! \! \! \! \! \! \! \frac{\pi}{4}-x\neq k\pi\\\! \! \! \! \! \! \! \! \! \! \! \! \! \! \! \! \! \! \! \! \! \! \! \! \!\frac{\pi}{4}+x\neq k\pi \\\frac{1}{2}(\cos 2x+\cos\frac{\pi}{2})\neq 0 \end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\neq \frac{\pi}{4}-k\pi\\x\neq-\frac{\pi}{4}+k\pi \\ x\neq\frac{\pi}{4}+k\frac{\pi}{2} \end{matrix}\right.\Leftrightarrow x\neq\frac{\pi}{4}+k\frac{\pi}{2}(k\in Z)\)

\(\tan\left ( \frac{\pi}{4}-x \right )\tan\left ( \frac{\pi}{4}+x \right )=\frac{\sin\left ( \frac{\pi}{4}-x \right )\sin\left ( \frac{\pi}{4}+x \right )}{\cos\left ( \frac{\pi}{4}-x \right )\cos\left ( \frac{\pi}{4}+x \right )}=\frac{\frac{1}{2}(\cos2x-\cos\frac{\pi}{2})}{\frac{1}{2}(\cos2x+\cos\frac{\pi}{2})}=1\)

\((1)\Leftrightarrow \sqrt{2}\sin2x=\sin x-\cos3x+2\cos2x\cos x\)
 

\(\Leftrightarrow \sqrt{2}\sin 2x=\sin x-\cos 3x+\cos x+\cos 3x\)

\(\Leftrightarrow \sqrt{2}\sin2x=\sqrt{2}\sin\left ( x+\frac{\pi}{4} \right )\)

\(\Leftrightarrow \sin2x=\sin\left ( x+\frac{\pi}{4} \right )\Leftrightarrow \bigg \lbrack \begin{matrix} 2x=x+\frac{\pi}{4}+k2\pi\\2x=\pi-(x+\frac{\pi}{4})+k2\pi \end{matrix}\Leftrightarrow \bigg \lbrack\begin{matrix} x=\frac{\pi}{4}+k2\pi\\x=\frac{\pi}{4}+k\frac{2\pi}{3} \end{matrix}\)

Kết hợp với điều kiện đã cho có nghiệm là:

\(x=\frac{11\pi}{12}+k2\pi,x=-\frac{5\pi}{12}+k2\pi(k\in Z)\)