4sin α . sin ( α - π/2 ) . sin ( 2α - π/2 ) + √3 cos 4α = 1

To solve the equation

\[
4 \sin \alpha \cdot \sin \left( \alpha - \frac{\pi}{2} \right) \cdot \sin \left( 2\alpha - \frac{\pi}{2} \right) + \sqrt{3} \cos 4\alpha = 1,
\]

we start by simplifying the sine terms.

1. **Use the sine subtraction identity**:
\[
\sin\left(\alpha - \frac{\pi}{2}\right) = -\cos \alpha,
\]
\[
\sin\left(2\alpha - \frac{\pi}{2}\right) = -\cos(2\alpha).
\]

Plugging these into the equation, it becomes:
\[
4 \sin \alpha \cdot (-\cos \alpha) \cdot (-\cos (2\alpha)) + \sqrt{3} \cos 4\alpha = 1.
\]

This simplifies to:
\[
4 \sin \alpha \cos \alpha \cos (2\alpha) + \sqrt{3} \cos 4\alpha = 1.
\]

2. **Use the double angle and product-to-sum identities**:
Recall that \( \sin(2\alpha) = 2 \sin \alpha \cos \alpha \) and \( \cos(2\alpha) = 2 \cos^2 \alpha - 1 \). We can rewrite \( 4 \sin \alpha \cos \alpha \) as \( 2 \sin (2\alpha) \):
\[
2 \sin(2\alpha) \cos(2\alpha) + \sqrt{3} \cos 4\alpha = 1.
\]

Now let's put this in another form using \( \sin(4\alpha) = 2 \sin(2\alpha) \cos(2\alpha) \):
\[
\sin(4\alpha) + \sqrt{3} \cos(4\alpha) = 1.
\]

3. **Convert to a single trigonometric function**:
Notice that:
\[
\sin(4\alpha) + \sqrt{3} \cos(4\alpha) = 1
\]
can be represented in a single sine function as:
\[
R \sin(4\alpha + \phi) = 1,
\]
where \( R = \sqrt{1^2 + (\sqrt{3})^2} = 2 \) and \( \phi = \tan^{-1}(\sqrt{3}/1) = \frac{\pi}{3} \).

Thus, the equation becomes:
\[
2 \sin(4\alpha + \frac{\pi}{3}) = 1,
\]
leading to:
\[
\sin(4\alpha + \frac{\pi}{3}) = \frac{1}{2}.
\]

4. **Solve for \( 4\alpha + \frac{\pi}{3} \)**:
The general solutions are:
\[
4\alpha + \frac{\pi}{3} = \frac{\pi}{6} + 2k\pi \quad \text{or} \quad 4\alpha + \frac{\pi}{3} = \frac{5\pi}{6} + 2k\pi \quad (k \in \mathbb{Z}).
\]

5. **Finding \( \alpha \)**:
From \( 4\alpha + \frac{\pi}{3} = \frac{\pi}{6} \):
\[
4\alpha = -\frac{\pi}{6} + \frac{\pi}{3} = \frac{\pi}{6} \implies \alpha = \frac{\pi}{24}.
\]

From \( 4\alpha + \frac{\pi}{3} = \frac{5\pi}{6} \):
\[
4\alpha = \frac{5\pi}{6} - \frac{\pi}{3} = \frac{5\pi}{6} - \frac{2\pi}{6} = \frac{3\pi}{6} = \frac{\pi}{2} \implies \alpha = \frac{\pi}{8}.
\]

Thus, the solutions for \( \alpha \) are \( \alpha = \frac{\pi}{24} + \frac{k\pi}{2} \) and \( \alpha = \frac{\pi}{8} + \frac{k\pi}{2} \) for integer \( k \).