Giải phương trình: 2cos^3x+cos2x+sinx=0

Hướng dẫn giải:

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

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

 \( \Leftrightarrow \left[ \begin{align}  & 1-\sin x=0 \\  & 1+2\sin x\cos x+2(\sin x+\cos x)=0 \\ \end{align} \right. \) \( \Leftrightarrow \left[ \begin{align}  & \sin x=1 \\  & {{(\sin x+\cos x)}^{2}}+2(\sin x+\cos x)=0 \\ \end{align} \right. \)

 \( \Leftrightarrow \left[ \begin{align}  & \sin x=1 \\  & (\sin x+\cos x)(\sin x+\cos x+2)=0 \\ \end{align} \right. \) \( \Leftrightarrow \left[ \begin{align}  & \sin x=1 \\  & \sin x+\cos x=0 \\  & \sin x+\cos x=-2\text{ }(\text{vô nghiệm }do\text{ }{{1}^{2}}+{{1}^{2}}<{{2}^{2}}) \\ \end{align} \right. \)

 \( \Leftrightarrow \left[ \begin{align}  & \sin x=1 \\  & \tan x=-1 \\ \end{align} \right. \) \( \Leftrightarrow \left[ \begin{align}  & x=\frac{\pi }{2}+k2\pi  \\  & x=-\frac{\pi }{4}+k2\pi  \\ \end{align} \right.,\text{ }k\in \mathbb{Z} \).