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Mathematics
Problem
If
2
sin
−
1
x
=
sin
−
1
(
2
x
1
−
x
2
)
2 \sin^{-1} x = \sin^{-1} (2x\sqrt{1-x^2})
2
sin
−
1
x
=
sin
−
1
(
2
x
1
−
x
2
)
, then
x
∈
x \in
x
∈
(a)
[
−
1
,
1
]
[-1, 1]
[
−
1
,
1
]
(b)
[
−
1
2
,
1
]
[-\frac{1}{\sqrt{2}}, 1]
[
−
2
1
,
1
]
(c)
[
−
1
2
,
1
2
]
[-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}]
[
−
2
1
,
2
1
]
(d) None of these
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