Giải phương trình: 4cos^3x+3√2sin2x=8cosx

Hướng dẫn giải:

Ta có: (*) \( \Leftrightarrow 4{{\cos }^{3}}x+6\sqrt{2}\sin x\cos x-8\cos x=0 \)

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

\(\Leftrightarrow \left[ \begin{align}  & \cos x=0 \\  & -2{{\sin }^{2}}x+3\sqrt{2}\sin x-2=0 \\ \end{align} \right.\)\(\Leftrightarrow \left[ \begin{align}  & \cos x=0 \\  & \sin x=\frac{\sqrt{2}}{2}=\sin \frac{\pi }{4} \\  & \sin x=\sqrt{2}\text{ }(\text{vô nghiệm }) \\ \end{align} \right.\)

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