Two numbers x and y in [-1, 1] satisfy sin⁻¹x + sin⁻¹y = 2π/3…

Step 1. Add the partner rules. $\left(\sin^{-1}x + \cos^{-1}x\right) + \left(\sin^{-1}y + \cos^{-1}y\right) = \dfrac{\pi}{2} + \dfrac{\pi}{2} = \pi$.

Step 2. The inverse cosines. $\cos^{-1}x + \cos^{-1}y = \pi - \dfrac{2\pi}{3} = \dfrac{\pi}{3} = 1.0472$.

Step 3. The case $x = y$. Then $2\sin^{-1}x = \dfrac{2\pi}{3}$, so $\sin^{-1}x = \dfrac{\pi}{3}$ and $x = \sin\dfrac{\pi}{3} = \dfrac{\sqrt{3}}{2} = 0.866$.