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Given that the inverse trigonometric functions take principal values only. Then, the number of real values of x which satisfy sin⁻¹(3x/5) + sin⁻¹(4x/5) = sin⁻¹ x is equal to:

1. 2
2. 1
3. 3
4. 0

Correct answer: 3

Solution

Taking sines reduces it to 25=3*sqrt(25-16x^2)+4*sqrt(25-9x^2), satisfied by x^2=1, and x=0 is trivially a solution. Checking principal values, x=-1,0,1 all work, so there are 3 solutions.

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