The value of $x$ satisfying the equation $(\sin^{-1}x)^3-(\cos^{-1}x)^3+(\sin^{-1}x)(\cos^{-1}x)(\sin^{-1}x-\cos^{-1}x)=\frac{\pi^3}{16}$ cannot be equal to:
Step-by-Step Solution
Key Concept: Use the constraint $\sin^{-1}x + \cos^{-1}x = \frac{\pi}{2}$ to reduce the equation to a single variable and factor using algebraic identities.
Let $a = \sin^{-1}x$ and $b = \cos^{-1}x$. We know that $a + b = \frac{\pi}{2}$, so $b = \frac{\pi}{2} - a$. The given equation becomes $a^3 - b^3 + ab(a-b) = \frac{\pi^3}{16}$. Using the factorization $a^3 - b^3 = (a-b)(a^2 + ab + b^2)$, we get $(a-b)(a^2 + ab + b^2 + ab) = (a-b)(a^2 + 2ab + b^2) = (a-b)(a+b)^2 = \frac{\pi^3}{16}$. Since $a + b = \frac{\pi}{2}$, we have $(a-b)\cdot\frac{\pi^2}{4} = \frac{\pi^3}{16}$, giving $a - b = \frac{\pi}{4}$. Combined with $a + b = \frac{\pi}{2}$, we get $a = \frac{3\pi}{8}$ and $b = \frac{\pi}{8}$. Therefore $x = \sin\frac{3\pi}{8} = \cos\frac{\pi}{8}$. The solution is $x = \cos\frac{\pi}{8}$.
Correct Answer: 1,2,4