Solve the equation: $\;$ $9^{x^2 - 1} - 36 \times 3^{x^2 - 3} + 3 = 0$
Given equation: $\;\;$ $9^{x^2 - 1} - 36 \times 3^{x^2 - 3} + 3 = 0$
i.e. $\;$ $\left[\left(3\right)^2\right]^{x^2 - 1} - 36 \times 3^{x^2} \times 3^{-3} + 3 = 0$
i.e. $\;$ $3^{2x^2 - 2} - \dfrac{36}{27} \times 3^{x^2} + 3 = 0$
i.e. $\;$ $3^{2x^2} \times 3^{-2} - \dfrac{4}{3} \times 3^{x^2} + 3 = 0$
i.e. $\;$ $\dfrac{1}{9} \times \left(3^{x^2}\right)^2 - \dfrac{4}{3} \times 3^{x^2} + 3 = 0$ $\;\;\; \cdots \; (1)$
Let $\;$ $3^{x^2} = p$ $\;\;\; \cdots \; (2)$
Then equation $(1)$ becomes
$\dfrac{1}{9} p^2 - \dfrac{4}{3} p + 3 = 0$
i.e. $\;$ $p^2 - 12p + 27 = 0$
i.e. $\;$ $\left(p - 3\right) \left(p - 9\right) = 0$
i.e. $\;$ $p = 3$ $\;$ or $\;$ $p = 9$
When $\;$ $p = 3$, $\;$ we have from equation $(2)$,
$3^{x^2} = 3 = 3^1$
$\implies$ $x^2 = 1$ $\implies$ $x = \pm 1$
When $\;$ $p = 9$, $\;$ we have from equation $(2)$,
$3^{x^2} = 9 = 3^2$
$\implies$ $x^2 = 2$ $\implies$ $x = \pm \sqrt{2}$
$\therefore \;$ The solution to the given equation is $\;\;$ $x = \left\{\pm \sqrt{2}, \; \pm 1 \right\}$