Step 1: Introduce a single variable.
The equation is $3^{2\sin^2x}+3^{2\cos^2x}=6$. Let $a=3^{2\sin^2x}$, a positive quantity.
Step 2: Express the second term through a.
Since $\cos^2x=1-\sin^2x$, we get $3^{2\cos^2x}=3^{2-2\sin^2x}=\frac{3^2}{3^{2\sin^2x}}=\frac{9}{a}$.
Step 3: Form one equation in a.
The equation becomes $a+\frac{9}{a}=6$. Multiply through by $a$: $a^2-6a+9=0$.
Step 4: Solve the perfect square.
This factors as $(a-3)^2=0$, so $a=3$, a unique value. (By AM-GM, $a+\frac9a\ge 6$, with equality only at $a=3$.)
Step 5: Translate back to x.
So $3^{2\sin^2x}=3$ means $2\sin^2x=1$, hence $\sin^2x=\frac12$ and $\sin x=\pm\frac{1}{\sqrt2}$.
Step 6: Count the solutions in the interval.
On $[-\pi,\pi]$ the values with $|\sin x|=\frac{1}{\sqrt2}$ are $x=\pm\frac{\pi}{4}$ and $x=\pm\frac{3\pi}{4}$, giving four solutions.
\[ \boxed{4} \]