Given that [tex]\(\tan(x) = \frac{7}{2}\)[/tex] and [tex]\(180^\circ \leq x \leq 270^\circ\)[/tex], which means [tex]\(x\)[/tex] is in the third quadrant where both sine and cosine are negative.
We start by using the identity of tangent:
[tex]\[
\tan(x) = \frac{\sin(x)}{\cos(x)} = \frac{7}{2}
\][/tex]
In the third quadrant, both [tex]\(\sin(x)\)[/tex] and [tex]\(\cos(x)\)[/tex] are negative. We can represent [tex]\(\sin(x)\)[/tex] as [tex]\(-7k\)[/tex] and [tex]\(\cos(x)\)[/tex] as [tex]\(-2k\)[/tex] for some positive constant [tex]\(k\)[/tex].
By using the Pythagorean identity:
[tex]\[
\sin^2(x) + \cos^2(x) = 1
\][/tex]
Substituting [tex]\(\sin(x) = -7k\)[/tex] and [tex]\(\cos(x) = -2k\)[/tex] into the identity:
[tex]\[
(-7k)^2 + (-2k)^2 = 1
\][/tex]
[tex]\[
49k^2 + 4k^2 = 1
\][/tex]
[tex]\[
53k^2 = 1
\][/tex]
Solving for [tex]\(k\)[/tex]:
[tex]\[
k^2 = \frac{1}{53}
\][/tex]
[tex]\[
k = \frac{1}{\sqrt{53}}
\][/tex]
Now, substituting [tex]\(k\)[/tex] back into [tex]\(\sin(x)\)[/tex] and [tex]\(\cos(x)\)[/tex]:
[tex]\[
\sin(x) = -7k = -\frac{7}{\sqrt{53}}
\][/tex]
[tex]\[
\cos(x) = -2k = -\frac{2}{\sqrt{53}}
\][/tex]
The double angle formula for sine is:
[tex]\[
\sin(2x) = 2 \sin(x) \cos(x)
\][/tex]
Substituting [tex]\(\sin(x)\)[/tex] and [tex]\(\cos(x)\)[/tex] into this formula:
[tex]\[
\sin(2x) = 2 \left(-\frac{7}{\sqrt{53}}\right) \left(-\frac{2}{\sqrt{53}}\right)
\][/tex]
[tex]\[
\sin(2x) = 2 \cdot \frac{14}{53}
\][/tex]
[tex]\[
\sin(2x) = \frac{28}{53}
\][/tex]
So, the value of [tex]\(\sin(2x)\)[/tex] is:
[tex]\[
\boxed{\frac{28}{53}}
\][/tex]
