Question 451563
2 cos^2 theta + 3 cos theta + 1 = 0
To solve, let x = cos{{{theta}}}
So we have 
2x^2 + 3x + 1 = 0
Solve using the quadratic formula:
x = (-3 +- sqrt(9 - 8))/4
This gives x = -1/2, x = -1
So we need to find the values of {{{theta}}} on the interval [0,2{{{pi}}}) which satisfy
cos{{{theta}}} = -1/2, -1
cos{{{pi}}} = -1, one of the zeros is {{{pi}}}
cos(2{{{pi}}}/3) = cos(4{{{pi}}}/3) = -1/2
So the zeros are: {{{theta}}} = (2/3){{{pi}}},{{{pi}}},(4/3){{{pi}}}
The graph is below:
{{{graph(400,400,-5,5,-2,6,2*(cos(x))^2+3*cos(x)+1)}}}