Solve cos⁻¹x + 2 sin⁻¹x = 2π/3 and sin⁻¹x + 2 cos⁻¹x = 3π/4…

Step 1. The first equation. $\dfrac{\pi}{2} - \sin^{-1}x + 2\sin^{-1}x = \dfrac{2\pi}{3}$, so $\sin^{-1}x = \dfrac{2\pi}{3} - \dfrac{\pi}{2} = \dfrac{\pi}{6}$ and $x = \dfrac{1}{2} = 0.5$.

Step 2. The second equation. $\sin^{-1}x + \pi - 2\sin^{-1}x = \dfrac{3\pi}{4}$, so $\sin^{-1}x = \pi - \dfrac{3\pi}{4} = \dfrac{\pi}{4}$ and $x = \dfrac{1}{\sqrt{2}} = 0.7071$.

Step 3. Check. Both values of $\sin^{-1}x$ are in $\left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right]$, and both $x$ are in $[-1, 1]$.