Simplify 60^ ecirc [(60^ ecirc + 2x) + (60^ ecirc - 2x)] given x = sqrt(t). Use the result from 7.2.1 and express in terms of t.

From 7.2.1, we found that $\sin(60^\circ + 2x) + \sin(60^\circ - 2x) = \sqrt{3} \cos 2x$.
The expression to evaluate is $\tan 60^\circ [\sin(60^\circ + 2x) + \sin(60^\circ - 2x)]$.
Substitute the result from 7.2.1:

$$\tan 60^\circ [\sqrt{3} \cos 2x]$$

We know that $\tan 60^\circ = \sqrt{3}$.
Substitute this value into the expression:

$$\sqrt{3} \cdot \sqrt{3} \cos 2x$$ $$= 3 \cos 2x$$

The identity is $\cos 2x = 2 \cos^2 x - 1$.
Substitute this into the expression:

$$3 (2 \cos^2 x - 1)$$

$$3 (2 (\sqrt{t})^2 - 1)$$ $$= 3 (2t - 1)$$

$$3(2t - 1) = 6t - 3$$

The value of the expression in terms of $t$ is:

$$6t - 3$$