Giải phương trình: (2sinx+1)(3cos4x+2sinx−4)+4cos^2x=3

Hướng dẫn giải:

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

 \( \Leftrightarrow (2\sin x+1)(3\cos 4x+2\sin x-4)+(1+2\sin x)(1-2\sin x)=0 \)

 \( \Leftrightarrow (2\sin x+1)\left[ 3\cos 4x+2\sin x-4+(1-2\sin x) \right]=0 \)

 \( \Leftrightarrow 3(\cos 4x-1)(2\sin x+1)=0\Leftrightarrow \left[ \begin{align}  & \cos 4x=1 \\  & \sin x=-\frac{1}{2}\sin \left( -\frac{\pi }{6} \right) \\ \end{align} \right. \)

 \( \Leftrightarrow \left[ \begin{align}  & 4x=k2\pi  \\  & x=-\frac{\pi }{6}+k2\pi  \\  & x=\frac{7\pi }{6}+k2\pi  \\ \end{align} \right. \)\(\Leftrightarrow \left[ \begin{align} & x=\frac{k\pi }{2} \\  & x=-\frac{\pi }{6}+k2\pi  \\  & x=\frac{7\pi }{6}+k2\pi  \\ \end{align} \right.,\text{ }k\in \mathbb{Z}\).