InterviewSolution
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InterviewSolution
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Solve `sin^(-1) (1 - x) - 2 sin^(-1) x = (pi)/(2)` |
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Answer» Given , `sin^(-1) (1 - x) - 2 sin^(-1) x = pi//2` Let `x sin y` Therefore, the given equation reduces sto `sin^(-1) (1 - sin y) - 2 sin^(-1) sin y= (pi)/(2)` or `sin^(-1) (1 - sin y) - 2 y = (pi)/(2)` or `sin^(-1) (1 - sin y) = (pi)/(2) + 2y` or `1 - sin y = sin ((pi)/(2) + 2y)` or `1 - sin y = cos 2 y` or `1 - cos 2y = sin y` or `2 sin^(2) y = sin y` or `2 sin^(2) y - sin y = 0` or `sin y (2 sin y - 1) = 0` `rArr sin y = 0 " or " 1//2` `rArr x = 0 " or " 1//2` But, when `x = (1)/(2)`, it can be observed that `L.H.S. = sin^(-1). (1 - (1)/(2)) - 2 sin ^(-1). (1)/(2)` `= sin^(-1). ((1)/(2)) - 2 sin^(-1). (1)/(2)` `= - sin^(-1). (1)/(2)` `= - (pi)/(6) != (pi)/(2)` Therefore, `x = (1)/(2)` is not the solution of the given equation Thus, `x = 0` |
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