Solution is {0,2π3,4π3}
Explanation:
2cos2x−cosx−1=0 can be written as
2cos2x−2cosx+cosx−1=0
or 2cosx(cosx−1)+1(cosx−1)=0
or (2cosx+1)(cosx−1)=0
∴ either 2cosx+1=0 i.e. cosx=−12 and in the interval [0,2π) x=2π3 or 4π3
or cosx−1= i.e. cosx=1 and in given interval x=0.
