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If $\sin^{- 1} (6 x) + \sin^{- 1} (6 \sqrt{3} x) = - \frac{π}{2}$ then the value of $x$ is

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112

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143

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detailed solution

Correct option is B

We have sin−1(6x)+sin−1(63x)=−π2⇒sin-16x+π2-cos-163x=-π2 ⇒π-cos-163x=-sin-16x ⇒cos-1-63x=sin-1-6x Let cos-1-63x=sin-1-6x=θ. ⇒cosθ=-63x and sinθ=-6x ∴sinθcosθ=-6x-63x ⇒tanθ=13 ⇒θ=π6 ∴sinθ=sinπ6=12=-6x ⇒x=-112

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If sin−1(6x)+sin−1(63x)=−π2then the value of x is

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If $\sin^{- 1} (6 x) + \sin^{- 1} (6 \sqrt{3} x) = - \frac{π}{2}$ then the value of $x$ is