Question

How do you solve `2 cos^2 (x) + cos(x) - 1 = 0` from [0,2pi]?

Answer 1

x=`pi/3, (5pi)/3`

Explanation:

Solving the quadratic expression in cos x, it would be `cosx = (-1+- sqrt(1+8))/2`

`cos x = (-1+-3)/2`

`cos x= 1, -2` cos x is not feasible, hence it is cos x= `1/2`, x=`pi/3, (5pi)/3`