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

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
$$5184x^4+1-144x^2=144x^2(1-108x^2)=144x^2-15552x^4$$
edited Mar 21, 2013