InterviewSolution
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InterviewSolution
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Solve `sin^(-1) (sin 6x) = x, x in [0,pi]` |
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Answer» `sin^(-1) (sin 6x) = x` `rArr sin 6 x = sin x` `rArr sin 6 x - sin x = 0` `rArr 2 sin.(5x)/(2) cos.(7x)/(2) = 0` `rArr (5x)/(2) n pi " or " (7x)/(2) = (2n + 1) (pi)/(2), n in Z` `rArr x = (2n pi)/(5) " or " x = (2n + 1) (pi)/(7)` `rArr x = (pi)/(7) , (3pi)/(7), (5pi)/(7), pi, 0, (2pi)/(5), (4pi)/(5)` But `x = (5 pi)/(7), pi, (4pi)/(5)` are not possible as any solution `[-(pi)/(2), (pi)/(2)]` |
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