To solve the expression \(\sin^2 x + \sin^2 58^\circ - \tan 40^\circ \cdot \tan 50^\circ\), we need to use some trigonometric identities and properties.

First, let's simplify the term \(\tan 40^\circ \cdot \tan 50^\circ\).

Using the identity for the tangent of complementary angles, we know that:
\[
\tan(90^\circ - \theta) = \cot(\theta)
\]
So,
\[
\tan 50^\circ = \cot 40^\circ
\]

Therefore,
\[
\tan 40^\circ \cdot \tan 50^\circ = \tan 40^\circ \cdot \cot 40^\circ = 1
\]

Now, let's consider the term \(\sin^2 x + \sin^2 58^\circ\). Note that \(\sin 58^\circ\) can be expressed in terms of \(\cos 32^\circ\) because \(\sin(90^\circ - \theta) = \cos(\theta)\). So,
\[
\sin 58^\circ = \cos 32^\circ
\]

Thus,
\[
\sin^2 58^\circ = \cos^2 32^\circ
\]

Now, let's rewrite the original expression:
\[
\sin^2 x + \cos^2 32^\circ - 1
\]

Using the Pythagorean identity, we know that:
\[
\sin^2 \theta + \cos^2 \theta = 1
\]

So, if \(x = 32^\circ\), then:
\[
\sin^2 32^\circ + \cos^2 32^\circ = 1
\]

Therefore, the expression simplifies to:
\[
1 - 1 = 0
\]

So, the value of the expression \(\sin^2 x + \sin^2 58^\circ - \tan 40^\circ \cdot \tan 50^\circ\) is \(0\).