To solve the trigonometric equation sin‌−1x=2sin‌−1a, we must consider the following: Given that sin‌−1x=2sin‌−1a, it implies: sin‌−1x∈[−‌
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2
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] This corresponds to: 2sin‌−1a∈[−‌
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2
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2
] From this, we can deduce: sin‌−1a∈[−‌
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4
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] Consequently, the values for a must satisfy: a∈[−sin‌‌