InterviewSolution
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InterviewSolution
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The trigonometric equation `sin^(-1)x=2sin^(-1)a`has a solution forall real values (b) `|a| |
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Answer» Let `alpha` a solution of `sin^(-1)x=2 sin^(-1) alpha` Then `sin^(-1) alpha =2 sin^(-1) alpha` `rarr -(pi)/(2) le 2 sin^(-1) alpha le (pi)/(2)` ` rarr -(pi)/(4) le sin^(-1) a le (pi)/(4)` `rarr sin((pi)/(4)) le alpha le sin(pi)/(4) rarr -(1)/sqrt(2) le alpha le (1)/sqrt(2) rarr |alpha| le (1)/sqrt(2)` |
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