solve for the interval of [0,2pi]

2cos^2(x)-7cos(x)-4=0    factor  this as :

(2cos(x) + 1 ) (cos(x) - 4)   = 0

Setting the second factor to 0, we have :

cos(x) - 4  = 0     add 4 to both sides

cos(x)  = 4     which has no solution

Setting the first factor to 0, we have

2cos(x)  + 1  = 0      subtract 1 from both sides

2cos(x)  =  -1        divide both side by 2

cos(x)  = -1/2      and this occurs at   2pi/3   and at 4pi/3    radians   on the requested interval