For $x \in [-1,1]$, if $4\sin^{-1}x + \cos^{-1}x = \pi$ then $x$ is equal to

Given: $4\sin^{-1}x + \cos^{-1}x = \pi$ where $x \in [-1,1]$

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The fundamental identity states:

$\sin^{-1}x + \cos^{-1}x = \frac{\pi}{2}$ for all $x \in [-1,1]$

This works because if an angle has sine = $x$, then its complementary angle has cosine = $x$, and they add up to $\frac{\pi}{2}$ radians.

From this identity:

$\cos^{-1}x = \frac{\pi}{2} - \sin^{-1}x$

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Substituting into the given equation:

$4\sin^{-1}x + \left(\frac{\pi}{2} - \sin^{-1}x\right) = \pi$

$4\sin^{-1}x - \sin^{-1}x + \frac{\pi}{2} = \pi$

$3\sin^{-1}x + \frac{\pi}{2} = \pi$

$3\sin^{-1}x = \pi - \frac{\pi}{2}$

$3\sin^{-1}x = \frac{\pi}{2}$

$\sin^{-1}x = \frac{\pi}{6}$

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Taking sine on both sides:

$x = \sin\left(\frac{\pi}{6}\right)$

$x = \frac{1}{2}$

Therefore, $x = \dfrac{1}{2}$