Find all real $x$ such that $0 < x < \pi $ and $\frac{8}{3 sin x - sin 3x} + 3 sin^2x \le 5$.
1. We start with the given inequality:
   \[
   \frac{8}{3 \sin x - \sin 3x} + 3 \sin^2 x \le 5
   \]
   
2. Using the triple angle identity for sine, we know:
   \[
   \sin 3x = 3 \sin x - 4 \sin^3 x
   \]
   
3. Substitute \(\sin 3x\) into the inequality:
   \[
   \frac{8}{3 \sin x - (3 \sin x - 4 \sin^3 x)} + 3 \sin^2 x \le 5
   \]
   Simplifying inside the denominator:
   \[
   \frac{8}{4 \sin^3 x} + 3 \sin^2 x \le 5
   \]
   
4. Let \(y = \sin x\). Then the inequality becomes:
   \[
   \frac{2}{y^3} + 3 y^2 \le 5
   \]
   
5. To solve this inequality, we first rearrange it:
   \[
   \frac{2}{y^3} + 3 y^2 - 5 \le 0
   \]
   
6. Multiply through by \(y^3\) (noting that \(y > 0\)):
   \[
   2 + 3 y^5 - 5 y^3 \le 0
   \]
   
7. Rearrange to form a polynomial inequality:
   \[
   3 y^5 - 5 y^3 + 2 \le 0
   \]
   
8. Let \(f(y) = 3 y^5 - 5 y^3 + 2\). We need to find the roots of \(f(y) = 0\) and determine where \(f(y) \le 0\).

9. Test \(y = 1\):
   \[
   f(1) = 3(1)^5 - 5(1)^3 + 2 = 3 - 5 + 2 = 0
   \]
   So, \(y = 1\) is a root.

10. Factor \(f(y)\) using \(y = 1\):
    \[
    f(y) = (y - 1)(3 y^4 + 3 y^3 - 2 y^2 - 2 y + 2)
    \]
    
11. Since \(y = 1\) is a root, we need to check the sign of \(f(y)\) in the interval \(0 < y \le 1\).

12. For \(0 < y < 1\), we need to check if \(f(y) \le 0\). Since \(f(y)\) is a polynomial of degree 5, and we know \(f(1) = 0\), we need to check the behavior of \(f(y)\) in the interval \(0 < y < 1\).

13. By testing values in the interval \(0 < y < 1\), we find that \(f(y) \le 0\) for \(0 < y \le 1\).

14. Therefore, the only solution in the interval \(0 < x < \pi\) is when \(\sin x = 1\), which occurs at:
    \[
    x = \frac{\pi}{2}
    \]

The final answer is \(\boxed{\frac{\pi}{2}}\)