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Solve the equation { sin }^{ -1 }6x+{ sin }^{ -1 }6 { 3 } x= {

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Solve the equation $s in^{- 1} 6 x + s in^{- 1} 63 x = \frac{- π}{2}$

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Text solutionVerified

From the given equation, we have

sin^{- 1} 6 x = \frac{- π}{2} - sin^{- 1} 63 x

\Rightarrow sin^{- 1} 6 x = - \frac{π}{2} - sin^{- 1} 6 3 x

\Rightarrow sin (sin^{- 1} 6 x) = sin (- \frac{π}{2} - sin^{- 1} 6 3 x)

\Rightarrow 6 x = - cos (sin^{- 1} 6 3 x)

\Rightarrow 6 x = - 1 - 108 x^{2}

. squaring both, we get

36 x^{2} = 1 - 108 x^{2}

\Rightarrow 144 x^{2} = 1

\Rightarrow x \pm \frac{1}{12}

Note that

x = \frac{1}{12}

is the only root of the equation as

x = \frac{1}{12}

does not satisfy it.

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Question Text	Solve the equation $s in^{- 1} 6 x + s in^{- 1} 63 x = \frac{- π}{2}$
Answer Type	Text solution:1
Upvotes	150