Giải phương trình: 4(sin^4x+cos^4x)+√3sin4x=2

Hướng dẫn giải:

Ta có: (*) \( \Leftrightarrow 4\left[ {{({{\sin }^{2}}x+{{\cos }^{2}}x)}^{2}}-2{{\sin }^{2}}x{{\cos }^{2}}x \right]+\sqrt{3}\sin 4x=2 \)

 \( \Leftrightarrow 4\left[ 1-\frac{1}{2}{{\sin }^{2}}2x \right]+\sqrt{3}\sin 4x=2\Leftrightarrow \cos 4x+\sqrt{3}\sin 4x=-1 \)

 \( \Leftrightarrow \frac{1}{2}\cos 4x+\frac{\sqrt{3}}{2}\sin 4x=-\frac{1}{2}\Leftrightarrow \cos \left( 4x-\frac{\pi }{3} \right)=\cos \frac{2\pi }{3}\Leftrightarrow \left[ \begin{align}  & 4x-\frac{\pi }{3}=\frac{2\pi }{3}+k2\pi  \\  & 4x-\frac{\pi }{3}=-\frac{2\pi }{3}+k2\pi  \\ \end{align} \right. \)

 \( \Leftrightarrow \left[ \begin{align}  & 4x=\pi +k2\pi  \\  & 4x=-\frac{\pi }{3}+k2\pi  \\ \end{align} \right. \) \( \Leftrightarrow \left[ \begin{align}  & x=\frac{\pi }{4}+\frac{k\pi }{2} \\  & x=-\frac{\pi }{12}+\frac{k\pi }{2} \\ \end{align} \right.,k\in \mathbb{Z} \).

Cách khác:

(*) \( \Leftrightarrow 2(1-{{\sin }^{2}}2x)+\sqrt{3}\sin 4x=0\Leftrightarrow 2{{\cos }^{2}}2x+2\sqrt{3}\sin 2x\cos 2x=0 \)

 \( \Leftrightarrow \cos 2x\left( \cos 2x+\sqrt{3}\sin 2x \right)=0\Leftrightarrow \left[ \begin{align}  & \cos 2x=0 \\  & \sqrt{3}\sin 2x=-\cos 2x \\ \end{align} \right. \)

\(\Leftrightarrow \left[ \begin{align} & \cos 2x=0 \\  & \tan 2x=-\frac{1}{\sqrt{3}}=\tan \left( -\frac{\pi }{6} \right) \\ \end{align} \right.\)\(\Leftrightarrow \left[ \begin{align}  & 2x=\frac{\pi }{2}+k2\pi  \\  & 2x=-\frac{\pi }{6}+k2\pi  \\ \end{align} \right.\)

\(\Leftrightarrow \left[ \begin{align}  & x=\frac{\pi }{4}+k\pi  \\  & x=-\frac{\pi }{12}+k\pi  \\ \end{align} \right.,\text{ }k\in \mathbb{Z}\).