TS EAMCET 10 May 2024 Shift 2 Solved Paper

To solve the trigonometric equation sin‌−1x=2sin‌−1a, we must consider the following:
Given that sin‌−1x=2sin‌−1a, it implies:
sin‌−1x∈[−‌

π

2

,‌

π

2

]
This corresponds to:
2sin‌−1a∈[−‌

π

2

,‌

π

2

]
From this, we can deduce:
sin‌−1a∈[−‌

π

4

,‌

π

4

]
Consequently, the values for a must satisfy:
a∈[−sin‌‌

π

4

,sin‌‌

π

4

]
This simplifies to:
a∈[−‌

1

√2

,‌

1

√2

]
Therefore, the condition for a is:
|a|≤‌

1

√2