To determine the values of 'x' that satisfy the equation 2cos−1x=sin‌−1(2x√1−x2), let's analyze the components: Let x=cos‌θ. This implies θ=cos−1x. The equation can be expressed as sin‌−1(2‌cos‌θ‌s‌i‌n‌θ), which simplifies to sin‌−1(sin‌2θ). Hence, 2θ=sin‌−1(sin‌2θ). For sin‌−1(sin‌2θ) to equal 2cos−1x, i.e., 2θ,2θ must lie within the range [0,π]. Given: θ=cos−1x, the valid range for θ is [0,π∕4]. Therefore, cos−1x∈[0,π∕4]. This implies: x lies within [‌
1
√2
,1] because cos(π∕4)=‌
1
√2
. Thus, the equation is satisfied for values of x in the interval [‌