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Solve the equation: $\sin^{-1}(6x) + \sin^{-1}(6\sqrt 3 x) =-\large \frac{\pi}{2}$

(A) $ x=-\large\frac{1}{2}$ (B) $ x=\large\frac{1}{12}$ (C) $ x=-\large\frac{1}{6}$ (D) $ x=-\large\frac{1}{12}$
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Toolbox:
  • $sin(-\theta)=-sin\theta$
  • $sin(\large\frac{\pi}{2}$$+\theta)=cos\theta$
  • $sin^{-1}x=cos^{-1}\sqrt{1-x^2}$
  • $sin^{-1}(sin\theta)=\theta$
  • $cos^{-1}(cos\theta)=\theta$
Given eqn. is written as $-sin^{-1}6x=\large\frac{\pi}{2}$ $+sin^{-1}(6\sqrt 3 x)$
Taking $sin$ on both the sides we get
$sin(-sin^{-1}6x)=sin(\large\frac{\pi}{2}$$+sin^{-1}6\sqrt 3 x)$
We know that $sin(-\theta)=-sin\theta$
$\Rightarrow\:-sin(sin^{-1}6x)=cos(sin^{-1}6\sqrt 3 x)$
But we know that $sin^{-1}x=cos^{-1}\sqrt{1-x^2}$
$\Rightarrow \:sin^{-1}6\sqrt 3 x=cos^{-1}\sqrt{1-(6\sqrt 3 x)^2}=cos^{-1}\sqrt{1-108x^2}$
and we also know that $sin^{-1}(sin\theta)=\theta$
$\Rightarrow\:-6x=cos(cos^{-1}\sqrt{1-108x^2}$
$\Rightarrow\: -6x=\sqrt{1-108x^2}$
Squaring on both the sides we get
$36x^2=1-108x^2$
$\Rightarrow\: 144x^2=1$
$\Rightarrow \:x^2=\large\frac{1}{144}$
$\Rightarrow x=\pm\large\frac{1}{12}$
But $x=\large\frac{1}{12}$ does not satisfy the eqn.
$\therefore\: x=-\large\frac{1}{12}$
answered May 24, 2013 by rvidyagovindarajan_1
 

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