Solve 2 (2x + 30) = 3 (2x + 30) for (2x)

We are given the equation:

$$2 \sin(2x + 30) = 3 \cos(2x + 30)$$

Divide both sides by $\cos(2x + 30)$:

$$\frac{2 \sin(2x + 30)}{\cos(2x + 30)} = \frac{3 \cos(2x + 30)}{\cos(2x + 30)}$$ $$2 \tan(2x + 30) = 3$$ $$\tan(2x + 30) = \frac{3}{2}$$

We know the co-function identity:

$$\sin(90^\circ - \theta) = \cos(\theta)$$

Therefore, for $\theta = 2x$:

$$\sin(90^\circ - 2x) = \cos(2x)$$

Our goal is to find the value of $\cos(2x)$.

Let $\alpha = 2x + 30$. We have $\tan(\alpha) = \frac{3}{2}$.
We can construct a right-angled triangle where the opposite side to angle $\alpha$ is 3 and the adjacent side is 2.
Using the Pythagorean theorem, the hypotenuse $h$ is:

$$h = \sqrt{3^2 + 2^2} = \sqrt{9 + 4} = \sqrt{13}$$

From this triangle, we can find $\sin(\alpha)$ and $\cos(\alpha)$:

$$\sin(\alpha) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{3}{\sqrt{13}}$$ $$\cos(\alpha) = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{2}{\sqrt{13}}$$

We know that $2x = \alpha - 30^\circ$.
Using the cosine subtraction formula $\cos(A - B) = \cos A \cos B + \sin A \sin B$:

$$\cos(2x) = \cos(\alpha - 30^\circ) = \cos(\alpha) \cos(30^\circ) + \sin(\alpha) \sin(30^\circ)$$

We know the exact values for $\cos(30^\circ)$ and $\sin(30^\circ)$:

$$\cos(30^\circ) = \frac{\sqrt{3}}{2}$$ $$\sin(30^\circ) = \frac{1}{2}$$

Substitute the values of $\sin(\alpha)$, $\cos(\alpha)$, $\sin(30^\circ)$, and $\cos(30^\circ)$:

$$\cos(2x) = \left(\frac{2}{\sqrt{13}}\right) \left(\frac{\sqrt{3}}{2}\right) + \left(\frac{3}{\sqrt{13}}\right) \left(\frac{1}{2}\right)$$ $$\cos(2x) = \frac{2\sqrt{3}}{2\sqrt{13}} + \frac{3}{2\sqrt{13}}$$ $$\cos(2x) = \frac{2\sqrt{3} + 3}{2\sqrt{13}}$$

$$\cos(2x) = \frac{2\sqrt{3} + 3}{2\sqrt{13}}$$

Using approximate values:

$$\sqrt{3} \approx 1.7320508$$ $$\sqrt{13} \approx 3.6055513$$ $$\cos(2x) \approx \frac{2(1.7320508) + 3}{2(3.6055513)}$$ $$\cos(2x) \approx \frac{3.4641016 + 3}{7.2111026}$$ $$\cos(2x) \approx \frac{6.4641016}{7.2111026}$$ $$\cos(2x) \approx 0.89641009$$

Rounding to 4 decimal places:

$$\cos(2x) \approx 0.8964$$

Since $\sin(90^\circ - 2x) = \cos(2x)$,

$$\sin(90^\circ - 2x) \approx 0.8964$$

The final answer is $\boxed{0.8964}$.