Question 389103
Use the trig identity,
{{{cos(2x)=2(cos(x))^2-1}}}
So then,
{{{3*cos(2x)=6(cos(x))^2-3}}}
and then,
{{{3*cos(2x) + cos(x) + 2 = 0}}}
{{{6(cos(x))^2-3+cos(x)+2=0}}}

{{{6*(cos(x))^2+cos(x)-1=0}}}
Use a substitution,
Let {{{u=cos(x)}}}
{{{6u^2+u-1=0}}}
You can factor this quadratic or solve it using the quadratic formula.
Once you have the solution for {{{u}}}, back substitute and find {{{x}}}.