Find all real solutions quadratic to the power quadratic equals 1

find all real solutions to
 * From NRICH,
 * $$\left(2-x^2\right)^{x^2-3 \sqrt{2} x+4}=1\,$$

Extension: What if x is permitted to be a complex number?
 * MathHelp replies,

First try
ab=1 when a=1 or b=0 (and a is nonzero), so the first try at a solution is to solve
 * $$x^2-3 \sqrt{2} x+4=0\,$$

and
 * $$2-x^2=1\,$$

and take the union of the results. The first equation gives us $$\sqrt{2}$$ and $$\sqrt{8}$$. The second equation gives us 1 and &minus;1. But then we need to throw out $$\sqrt{2}$$ because that solution gives us 00, which is undefined. So the results of the first try are:
 * $$\sqrt{8}, \ 1, \ \mbox{and} \ 1. \,$$

Second try
Noting that (-1)2=1 and that $$e^{2 i \pi k}=1\,$$ for all integers, k, there are certainly other solutions to ab=1. For all I know at this point, there may be real solutions in x to the pair of equations
 * $$x^2-3 \sqrt{2} x+4=a\,$$

and
 * $$2-x^2=b,\,$$

where a and b are complex solutions to ab=1.

So I would like to start by characterizing solutions to ab=1. First, the equation is solved whenever a=1, regardless of b. If a is not equal to 1, then,


 * ab = eb ln(a) = $$\mathrm{e}^{b \ln{a} - 2 i \pi k}=1\,$$

and so the complete solution to ab=1 is
 * $$a=1,\,$$
 * $$a \ne 0, \ a \ne 1, \ b = \frac{2 i \pi k}{\ln(a)}\,$$

The second solution can also be expressed as
 * $$a = e^{\frac{2 i \pi k}{b} }\,$$

External references

 * NRICH weekly challenge 2: Quad Solve
 * Wolfram Alpha's solution