Solving $\sec^2(x)-2\sqrt{2} \sec(x)+2=0$

Question

What are the values of $x$ in degrees when: $$ \sec^2(x)-2\sqrt{2} \sec(x)+2=0 $$ I believe I should use derivatives of $\sec(x)$ but am struggling. Any help would be appreciated.

Answer 1

Hint:

the equation is:

$$ \left(\sec x -\sqrt{2}\right)^2=0 $$

Answer 2

Hint:

This is a quadratic, so you can use the substitution $u = \sec x$, which yields

$$u^2-2\sqrt2u+2 = 0$$

which so happens to be a perfect square trinomial which can be factored, yielding

$$(u-\sqrt 2)^2 = 0$$