In the closed interval [0,2 π], find the solution set of the equation | 2 (cosx) ^ 2-1 | = SiNx

In the closed interval [0,2 π], find the solution set of the equation | 2 (cosx) ^ 2-1 | = SiNx

|2(cosx)^2-1|=sinx
|1-2(sinx)^2|=sinx
1-2(sinx)^2=±sinx
(i)1-2(sinx)^2=sinx
2(sinx)^2+sinx-1=0
SiNx = - 1 or 1 / 2
So x = 3 π / 2 or π / 6 or 5 π / 6
(ii)1-2(sinx)^2=-sinx
2(sinx)^2-sinx-1=0
SiNx = 1 or - 1 / 2
So x = π / 2 or 7 π / 6 or 11 π / 6
So the solution set is {π / 6, π / 2,5 π / 6,7 π / 6,3 π / 2,11 π / 6}