a. ` cos3x-cos2x-cosx+1=0`
`<=>4cos^3x-3cosx-2cos^2x+1-cosx+1=0`
`<=>4cos^3x-2cos^2x-4cosx+2=0`
`<=>`$\left[\begin{matrix} \cos x=-1\\ \cos x=1\\\cos x =\dfrac{1}{2}\end{matrix}\right.$
`<=>` $\left[\begin{matrix} \sin x=0\\ \cos x=\dfrac{1}{2}\end{matrix}\right.$
`<=>` $\left[\begin{matrix} x=k\pi\\ x=±\dfrac{\pi}{3}+k2\pi\end{matrix}\right.$`(kinZZ)`
b. `cos4x-4cos2x+cosx-6=0`
Ta có:
`+`` cos4x=2cos^2 2x-1=2(2cos^2x-1)^2-1`
`=2(4cos^4x-4cos^2x+1)-1=8cos^4-8cos^2x+1`
`+``cos2x=2cos^2x-1`
Đặt `t=cosx` `(|t|leq1)`
Phương trình: `8t^4-8t^2+1-4(2t^2-1)+9t-6=0`
`<=>8t^4-16t^2+9t-1=0`
`<=>(t-1)(8t^3+8t^2-8t+1)=0`
`<=>`$\left[\begin{matrix} t=1\\ t=\dfrac{1}{2}\\t=\dfrac{-3+\sqrt{13}}{4}&(tm)\\t=\dfrac{-3-\sqrt{13}}{4}&(loại)&\end{matrix}\right.$
`<=>`$\left[\begin{matrix} x=k2\pi\\ x=-\dfrac{\pi}{3}+k2\pi\\x=\dfrac{\pi}{3}+k2\pi\\x=±\arccos(\dfrac{-3+\sqrt{13}}{4})+k2\pi&\end{matrix}\right.$`(kinZZ)`
