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# Solve the equation $sin^{-1}6x+sin^{-1}6\sqrt 3 x = -\frac{\pi}{2}.$

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Toolbox:
• $sin^{-1}x+sin^{-1}y=sin^{-1}\: x\sqrt{1-y^2}+y\sqrt{1-x^2}$
• $sin(-\frac{\pi}{2})=-1$
By taking 6x in the place of x and $6\sqrt3x$ in the place of y in the formula of $sin^{-1}x+sin^{-1}y$ we get
$sin^{-1}6x+sin^{-1}6\sqrt3x=$
$sin^{-1} \bigg[ 6x \sqrt{1-108x^2}+6\sqrt 3 x\sqrt{1-36x^2} \bigg]$$= -\frac{\pi}{2}$
Taking sin on both the sides
$\Rightarrow\: 6x\sqrt{1-108x^2}+6\sqrt 3 x\sqrt{1-36x^2}=sin(-\frac{\pi}{2})=-1$
We know that $sin(-\frac{\pi}{2})=-1$
$\Rightarrow 6\sqrt 3 x \sqrt{1-36x^2}=-1-6x\sqrt{1-108x^2}$
Squaring on both the sides we get
$\Rightarrow 108x^2(1-36x^2)=1+36x^2(1-108x^2)+12x\sqrt{1-108x^2}$
$\Rightarrow\: 72x^2-1=12x\sqrt{1-108x^2}$
Again squaring both the sides we get
$5184x^4+1-144x^2=144x^2(1-108x^2)=144x^2-15552x^4$
answered Feb 28, 2013
edited Mar 21, 2013