Giải phương trình: \( (2\sin x+1)(3\cos 4x+2\sin x-4)+4{{\cos }^{2}}x=3 \) (*)
Hướng dẫn giải:
Ta có: (*) \( \Leftrightarrow (2\sin x+1)(3\cos 4x+2\sin x-4)+4(1-{{\sin }^{2}}x)-3=0 \)
\( \Leftrightarrow (2\sin x+1)(3\cos 4x+2\sin x-4)+(1+2\sin x)(1-2\sin x)=0 \)
\( \Leftrightarrow (2\sin x+1)\left[ 3\cos 4x+2\sin x-4+(1-2\sin x) \right]=0 \)
\( \Leftrightarrow 3(\cos 4x-1)(2\sin x+1)=0\Leftrightarrow \left[ \begin{align} & \cos 4x=1 \\ & \sin x=-\frac{1}{2}\sin \left( -\frac{\pi }{6} \right) \\ \end{align} \right. \)
\( \Leftrightarrow \left[ \begin{align} & 4x=k2\pi \\ & x=-\frac{\pi }{6}+k2\pi \\ & x=\frac{7\pi }{6}+k2\pi \\ \end{align} \right. \)\(\Leftrightarrow \left[ \begin{align} & x=\frac{k\pi }{2} \\ & x=-\frac{\pi }{6}+k2\pi \\ & x=\frac{7\pi }{6}+k2\pi \\ \end{align} \right.,\text{ }k\in \mathbb{Z}\).
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