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प्रश्न
Solve the equation `sin^-1 6x + sin^-1 6sqrt(3)x = - pi/2`
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उत्तर
From the given equation
we have `sin^-1 6x = - pi/2 - sin^-1 6sqrt(3)x`
⇒ `sin(sin^-1 6x) = sin(- pi/2 - sin^-1 6sqrt(3)x)`
⇒ 6x = `- cos(sin^-1 6sqrt(3)x)`
⇒ 6x = `-sqrt(1 - 108x^2)`.
Squaring, we get
`36x^2= 1 - 108x^2`
⇒ 144x2 = 1
⇒ x = `+- 1/12`
Note that x = `- 1/12` is the only root of the equation as x = `1/12` does not satisfy it.
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