Find all real $x$ in the interval $[0, 2\pi)$ such that
\[27 \cdot 3^{3 \sin x} = 9^{\cos^2 x}.\]
1. Start with the given equation:
   \[
   27 \cdot 3^{3 \sin x} = 9^{\cos^2 x}
   \]

2. Rewrite the constants as powers of 3:
   \[
   27 = 3^3 \quad \text{and} \quad 9 = 3^2
   \]
   Thus, the equation becomes:
   \[
   3^3 \cdot 3^{3 \sin x} = (3^2)^{\cos^2 x}
   \]

3. Simplify the left-hand side using the properties of exponents:
   \[
   3^{3 + 3 \sin x} = 3^{2 \cos^2 x}
   \]

4. Since the bases are the same, set the exponents equal to each other:
   \[
   3 + 3 \sin x = 2 \cos^2 x
   \]

5. Use the Pythagorean identity \(\cos^2 x = 1 - \sin^2 x\):
   \[
   3 + 3 \sin x = 2 (1 - \sin^2 x)
   \]

6. Distribute and rearrange the equation:
   \[
   3 + 3 \sin x = 2 - 2 \sin^2 x
   \]
   \[
   2 \sin^2 x + 3 \sin x + 1 = 0
   \]

7. Solve the quadratic equation \(2 \sin^2 x + 3 \sin x + 1 = 0\). Let \(u = \sin x\):
   \[
   2u^2 + 3u + 1 = 0
   \]

8. Factor the quadratic equation:
   \[
   (2u + 1)(u + 1) = 0
   \]

9. Solve for \(u\):
   \[
   2u + 1 = 0 \quad \Rightarrow \quad u = -\frac{1}{2}
   \]
   \[
   u + 1 = 0 \quad \Rightarrow \quad u = -1
   \]

10. Substitute back \(u = \sin x\):
    \[
    \sin x = -\frac{1}{2} \quad \text{or} \quad \sin x = -1
    \]

11. Find the values of \(x\) in the interval \([0, 2\pi)\) that satisfy these conditions:
    - For \(\sin x = -\frac{1}{2}\):
      \[
      x = \frac{7\pi}{6}, \frac{11\pi}{6}
      \]
    - For \(\sin x = -1\):
      \[
      x = \frac{3\pi}{2}
      \]

12. Combine all solutions:
    \[
    x \in \left\{ \frac{7\pi}{6}, \frac{3\pi}{2}, \frac{11\pi}{6} \right\}
    \]

The final answer is \(\boxed{x \in \left\{ \frac{7\pi}{6}, \frac{3\pi}{2}, \frac{11\pi}{6} \right\}}\)