Given A = 2 sin² x – cos 2x

[ using cos 2x = 1 – 2 sin² x ]

so A = 2 sin² x – cos 2x = 2 sin² x –[ 1 – 2 sin² x]

= 2 sin² x -1 + 2 sin² x]

= 4 sin² x – 1

Now A = 2 sin² x – cos 2x = 4 sin² x – 1

As we know sin x lies between -1 and 1

-1 ≤ sin x ≤ 1

0 ≤ sin²x ≤ 1

Multiplying the inequality by 4

0 ≤ 4 sin² x ≤ 4

Subtracting 1 from the inequality

-1 ≤ (4 sin² x – 1) ≤ 3

From the above inequation, we can say that

A = (4 sin² x – 1) belongs to the closed interval [-1,3]

Hence the answer is A.