Simplify the expression: ^-1((3)/(5) x + (4)/(5) x)

The expression inside the inverse cosine is in the form $A \cos x + B \sin x$. We can relate this to the cosine addition formula $\cos(A-B) = \cos A \cos B + \sin A \sin B$.
Let's try to find an angle $\alpha$ such that $\cos \alpha = \frac{3}{5}$ and $\sin \alpha = \frac{4}{5}$.
We check if $\cos^2 \alpha + \sin^2 \alpha = 1$:

$$\left(\frac{3}{5}\right)^2 + \left(\frac{4}{5}\right)^2 = \frac{9}{25} + \frac{16}{25} = \frac{25}{25} = 1$$

Since the sum of squares is 1, such an angle $\alpha$ exists.

Let $\alpha = \cos^{-1}\left(\frac{3}{5}\right)$. This implies $\cos \alpha = \frac{3}{5}$.
From $\cos^2 \alpha + \sin^2 \alpha = 1$, we have $\sin \alpha = \sqrt{1 - \cos^2 \alpha} = \sqrt{1 - \left(\frac{3}{5}\right)^2} = \sqrt{1 - \frac{9}{25}} = \sqrt{\frac{16}{25}} = \frac{4}{5}$.
So, we can rewrite the expression as:

$$\cos^{-1}\left(\cos \alpha \cos x + \sin \alpha \sin x\right)$$

Using the identity $\cos(A-B) = \cos A \cos B + \sin A \sin B$, we can simplify the argument of the inverse cosine:

$$\cos^{-1}\left(\cos(\alpha - x)\right)$$

Since $\cos^{-1}(\cos \theta) = \theta$ for $\theta$ in the principal range, we get:

$$\alpha - x$$

Substitute $\alpha = \cos^{-1}\left(\frac{3}{5}\right)$ back into the expression:

$$\cos^{-1}\left(\frac{3}{5}\right) - x$$