Giải phương trình: 4sin^3x.cos3x+4cos^3x.sin3x+3√3cos4x=3

Hướng dẫn giải:

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

 \( \Leftrightarrow -12{{\sin }^{3}}xcosx+12\sin x{{\cos }^{3}}x+3\sqrt{3}\cos 4x=3 \)

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

 \( \Leftrightarrow 2\sin 2x.\cos 2x+\sqrt{3}\cos 4x=1\Leftrightarrow \sin 4x+\sqrt{3}\cos 4x=1 \)

 \( \Leftrightarrow \frac{1}{2}\sin 4x+\frac{\sqrt{3}}{2}\cos 4x=\frac{1}{2}\Leftrightarrow \sin \left( 4x+\frac{\pi }{3} \right)=\sin \frac{\pi }{6} \)

 \( \Leftrightarrow \left[ \begin{align}  & 4x+\frac{\pi }{3}=\frac{\pi }{6}+k2\pi  \\  & 4x+\frac{\pi }{3}=\pi -\frac{\pi }{6}+k2\pi  \\ \end{align} \right. \) \( \Leftrightarrow \left[ \begin{align}  & x=-\frac{\pi }{24}+\frac{k\pi }{2} \\  & x=\frac{\pi }{8}+\frac{k\pi }{2} \\ \end{align} \right.,\text{ }k\in \mathbb{Z} \).