Given the equation \(\sin^{-1}(4x^4 + x^2) = \frac{1}{6}\pi\), we first take the sine of both sides to get:

\(4x^4 + x^2 = \sin\left(\frac{1}{6}\pi\right)\).

Since \(\sin\left(\frac{1}{6}\pi\right) = \frac{1}{2}\), we have:

\(4x^4 + x^2 = \frac{1}{2}\).

Let \(y = x^2\), then the equation becomes:

\(4y^2 + y - \frac{1}{2} = 0\).

This is a quadratic equation in \(y\). Solving for \(y\) using the quadratic formula:

\(y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 4\), \(b = 1\), \(c = -\frac{1}{2}\).

\(y = \frac{-1 \pm \sqrt{1 + 8}}{8}\).

\(y = \frac{-1 \pm 3}{8}\).

This gives \(y = \frac{1}{4}\) or \(y = -\frac{1}{2}\).

Since \(y = x^2\) and \(y\) must be non-negative, we have \(x^2 = \frac{1}{4}\).

Thus, \(x = \pm \frac{1}{2}\).