Solve 2cos^2x + cosx − 1 = 0 for x over the interval [0, 2 pi ).

2cos²x + cosx − 1 = 0
(cosx+1)(2cosx-1)=0
so cosx+1=0
or 2cosx-1=0
1) cosx+1=0
cosx=-1
x=π+2πk, k∈Z

2) 2cosx-1=0
2cosx=1
cosx=0.5
x=+-, k∈Z

find value for the interval [0, 2 pi )

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This could also be written as, where 



This would factorize to give:

 

So we can factorize our original expression:



We can then solve for  as we would with a normal quadratic:



And also:



So our values for  are:



As:

Our final solutions for  are: