<s>[INST] Value of $x$ in $\sin^{-1}(x)+\sin^{-1}(1-x)=\cos^{-1}(x)$ How can we find the value of $x$ in $\sin^{-1}(x)+\sin^{-1}(1-x)=\cos^{-1}(x)$? 
Note that $\sin^{-1}$ is the inverse sine function.
I'm asking for the solution $x$ for this equation.
Please work out the solution.
 [/INST] $\bf{My\; Solution::}$ Given $$\sin^{-1}(x)+\sin^{-1}(1-x) = \cos^{-1}(x)$$
$$\displaystyle \sin^{-1}(1-x) = \cos^{-1}(x)-\sin^{-1}(x)\Rightarrow (1-x) = \sin \left[\cos^{-1}(x)-\sin^{-1}(x)\right]$$
Using $$\sin^{-1}(x)+\cos^{-1}(x) = \frac{\pi}{2}$$
So $$\displaystyle (1-x) = \sin\left[\frac{\pi}{2}-2\sin^{-1}(x)\right]=\cos(2\sin^{-1}(x)) = \cos(\cos^{-1}(1-2x^2))$$
Using $$2\sin^{-1}(x)=\cos^{-1}(1-2x^2)$$
So $$\displaystyle (1-x) = (1-2x^2)\Rightarrow 2x^2-x = 0$$
So $$\displaystyle 2x^2-x = 0\Rightarrow x = 0\;\;,x = \frac{1}{2}$$
Now Put into Original Equation we get $\displaystyle x = \frac{1}{2}$ and 
$x=0$ satisfy the Given equation.
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