To solve the equation \(\sin(2x + \pi/2) = \sinh(x - \pi/3)\), we need to consider the properties of the sine and hyperbolic sine functions.

1. **Simplify the Sine Expression:**
The identity for sine is \(\sin(a + b) = \sin a \cos b + \cos a \sin b\). Applying it:
\[
\sin(2x + \pi/2) = \sin 2x \cos(\pi/2) + \cos 2x \sin(\pi/2).
\]
Since \(\cos(\pi/2) = 0\) and \(\sin(\pi/2) = 1\), it follows that:
\[
\sin(2x + \pi/2) = \cos 2x.
\]

2. **Rewriting the Equation:**
The equation becomes:
\[
\cos 2x = \sinh(x - \pi/3).
\]

3. **Consider the Domain and Range:**
The function \(\cos 2x\) ranges from -1 to 1 for real values of \(x\).
The function \(\sinh(x - \pi/3)\) can output any real number since \(\sinh\) is not bounded.

Thus, we are looking for the points where \(\cos 2x\) (bounded function) equals \(\sinh(x - \pi/3)\) (continuous and unbounded function). For real solutions, both sides' domains must match, suggesting approximate solutions intersect for specific \(x\) values.

4. **Finding Solutions:**
Without algebraically solving directly (sometimes transcendental equations like this may not have closed-form solutions), numerical or graphical methods such as plotting \( \cos 2x \) and \(\sinh(x - \pi/3)\) would indicate where two sides intersect if solutions exist:

```python
import numpy as np
import matplotlib.pyplot as plt

x_values = np.linspace(-10, 10, 1000)
cos2x = np.cos(2 * x_values)
sinh_x_pi_3 = np.sinh(x_values - np.pi/3)

plt.plot(x_values, cos2x, label='cos(2x)')
plt.plot(x_values, sinh_x_pi_3, label='sinh(x - π/3)')
plt.axhline(0, color='gray', lw=0.5)
plt.axvline(0, color='gray', lw=0.5)
plt.legend()
plt.show()
```

5. **Conclusion:**
Given the bounded nature of \(\cos 2x\) compared to the behavior of \(\sinh(x-\pi/3)\), solutions may appear where both graphs intersect within -1 to 1 for specific \(x\) values numerically or graphically identified.