Giải phương trình: 1+sin^32x+cos^32x=1/2sin4x

Hướng dẫn giải:

Ta có: (*) \( \Leftrightarrow 1+(\sin 2x+\cos 2x)(1-\sin 2x\cos 2x)=\frac{1}{2}\sin 4x \)

 \( \Leftrightarrow 1-\frac{1}{2}\sin 4x+(\sin 2x+\cos 2x)\left( 1-\frac{1}{2}\sin 4x \right)=0\Leftrightarrow \left( 1-\frac{1}{2}\sin 4x \right)(1+\sin 2x+\cos 2x)=0 \)

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

 \( \Leftrightarrow \left[ \begin{align}  & 2x+\frac{\pi }{4}=-\frac{\pi }{4}+k2\pi  \\  & 2x+\frac{\pi }{4}=\frac{5\pi }{4}+k2\pi  \\ \end{align} \right. \) \( \Leftrightarrow \left[ \begin{align}  & x=-\frac{\pi }{4}+k\pi  \\  & x=\frac{\pi }{2}+k\pi  \\ \end{align} \right.,k\in \mathbb{Z} \).