If sin−1⁡2x1−x2−2sin−1⁡x=0, then x belongs to the interval

Let sin−1⁡x=θ. Then, x=sin⁡θ and 1−x2=cos⁡θNow,     sin−1⁡2x1−x2=sin−1⁡(sin⁡2θ)=2θ, if −π2≤20≤π2=2sin−1⁡x, if −π4≤θ≤π4 i.e. if −12≤x≤12∴ sin−1⁡2x1−x2−2sin−1⁡x=0, if -12≤x≤12