$f\left(x\right)\=\left(2 \cos \left(x\right)\+1\right) \left(3 \cos \left(x\right)-2\right)$ 

$2 \cos \left(x\right)\+1\=0$     and     $3 \cos \left(x\right)-2\=0$ 

Solution by Context Panel

Loading Student:-Calculus1

$6 {\cos }^2\left(x\right)-\cos \left(x\right)-2\=0$

$6{\cos \left(x\right)}^2-\cos \left(x\right)-2\=0$

$\stackrel{\text{roots}}{\to }$

$\left[\arccos \left(\frac{2}{3}\right)\,\frac{2}{3}\mathrm{\π}\,\frac{4}{3}\mathrm{\π}\,2\mathrm{\π}-\arccos \left(\frac{2}{3}\right)\right]$

$\stackrel{\text{at 5 digits}}{\to }$

$\left[0.84107\,2.0944\,4.1887\,5.4421\right]$

Solution from first principles

$6 {\cos }^2\left(x\right)-\cos \left(x\right)-2$

$6{\cos \left(x\right)}^2-\cos \left(x\right)-2$

$\stackrel{\text{factor}}{\=}$

$\left(2\cos \left(x\right)\+1\right)\left(3\cos \left(x\right)-2\right)$

$2\cos \left(x\right)\+1\=0$

$2\cos \left(x\right)\+1\=0$

$\stackrel{\text{roots}}{\to }$

$\left[\frac{2}{3}\mathrm{\π}\,\frac{4}{3}\mathrm{\π}\right]$

$\stackrel{\text{at 5 digits}}{\to }$

$\left[2.0944\,4.1887\right]$

$3\cos \left(x\right)-2\=0$

$3\cos \left(x\right)-2\=0$

$\stackrel{\text{roots}}{\to }$

$\left[\arccos \left(\frac{2}{3}\right)\,2\mathrm{\π}-\arccos \left(\frac{2}{3}\right)\right]$

$\stackrel{\text{at 5 digits}}{\to }$

$\left[0.84107\,5.4421\right]$

Loading Student:-Calculus1

$\mathrm{q\_\_1}≔6 {\cos }^2\left(x\right)-\cos \left(x\right)-2\=0\:$

$\mathrm{q\_\_2}≔\mathrm{Roots}\left(\mathrm{q\_\_1}\,x\=0..2 \mathrm{\π}\right)$

$\left[\arccos \left(\frac{2}{3}\right)\,\frac{2}{3}\mathrm{\π}\,\frac{4}{3}\mathrm{\π}\,2\mathrm{\π}-\arccos \left(\frac{2}{3}\right)\right]$

$\mathrm{evalf}\left(\mathrm{q\_\_2}\,5\right)$

$\left[0.84107\,2.0944\,4.1887\,5.4421\right]$

Solution from first principles

$\mathrm{q\_\_2}≔\mathrm{factor}\left(\mathrm{q\_\_1}\right)$

$\left(2\cos \left(x\right)\+1\right)\left(3\cos \left(x\right)-2\right)\=0$

$\mathrm{q\_\_3}≔\mathrm{lhs}\left(\mathrm{q\_\_2}\right)$

$\left(2\cos \left(x\right)\+1\right)\left(3\cos \left(x\right)-2\right)$

Apply the Roots and evalf commands to each factor, extracting factors with the op command

$\mathrm{q\_\_4}≔\mathrm{op}\left(1\,\mathrm{q\_\_3}\right)$

$2\cos \left(x\right)\+1$

$\mathrm{q\_\_5}≔\mathrm{Roots}\left(\mathrm{q\_\_4}\,x\=0..2 \mathrm{\π}\right)$

$\left[\frac{2}{3}\mathrm{\π}\,\frac{4}{3}\mathrm{\π}\right]$

$\mathrm{evalf}\left(\mathrm{q\_\_5}\,5\right)$

$\left[2.0944\,4.1887\right]$

$\mathrm{q\_\_6}≔\mathrm{op}\left(2\,\mathrm{q\_\_3}\right)$

$3\cos \left(x\right)-2$

$\mathrm{q\_\_7}≔\mathrm{Roots}\left(\mathrm{q\_\_6}\,x\=0..2 \mathrm{\π}\right)$

$\left[\arccos \left(\frac{2}{3}\right)\,2\mathrm{\π}-\arccos \left(\frac{2}{3}\right)\right]$

$\mathrm{evalf}\left(\mathrm{q\_\_7}\,5\right)$

$\left[0.84107\,5.4421\right]$