Generalizing Solutions for Trigonometric functions (Secant) I have been working on the problem: $3\sec^2(2x)-5=1$
Where you have to solve for x and then enter the generalized solutions in ascending order. This is using the Acrobatiq platform, but I am uncertain how to format the answer.
Thus far I performed the following:

*

*$3\sec^2(2x)-5=1$ (+5 both sides)

*$3\sec^2(2x)=6$

*$\sec^2(2x)=2$ (Divide 3 both sides)

*$\sqrt{\sec^2(2x)}=\sqrt2$ (Square both sides)

*$\sec(2x)=\pm \sqrt 2$
From $\pm \sqrt 2$ I get the points $\frac{\pi}{4}$ and $\frac{5 \pi}{4}$
Generalizing the solution I get $2x=\frac{\pi}{4}+2\pi k$ which I then obtain the solution: $\frac{9 \pi}{8}$ .
For $\frac{5 \pi}{4}$ I get $2x=\frac{5 \pi}{4}+2 \pi k$ = $\frac{13 \pi}{8}$
Dividing both solutions by 2 I get: ${\frac{9 \pi}{4}, \frac{13 \pi}{4}}$
(The question is stating that it wants the answer for $2\pi k$ though there are other solutions i.e. $\frac{\pi}{8}$)
When entering these answers for $x$ its incorrect.. am I performing something incorrect? Or is there specific formatting to the Acrobatiq platform? (If anyone is familiar)
 A: (1) Your general sub-solutions were both slightly incorrect, and (2) you'd made arithmetic mistakes. So, the correction is:
$\quad2x=\pm\frac\pi4+2\pi k \implies x=\frac\pi8(8k\pm1) \implies x=\pm\frac\pi8,\pm\frac{7\pi}8,\pm\frac{9\pi}8,\ldots$
$\quad2x=\pm\frac{5\pi}4+2\pi k \implies x=\frac\pi8(8k\pm5) \implies x=\pm\frac{3\pi}8,\pm\frac{5\pi}8,\pm\frac{11\pi}8,\pm\frac{13\pi}8,\ldots$
And notice that the second sub-solution would have been nicer had you started out with the $\sec(2x)$ root that is nearest 0 (i.e., $-\frac{3\pi}4$ instead of $\frac{5\pi}4$). Then it would've been rewritten more simply as $x=\frac\pi8(8k\pm3).$
Also, the two sub-solutions can be combined to give a final general solution $$x=\frac\pi8(2n+1).$$
P.S. Incidentally, it's worth checking out this thread, which explains how you're misusing the “=” sign.
A: I think you should use the definition of $\sec$ and congruences to solve it in a simple way:
$$\sec^22x=2\iff \cos^2 2x=\frac12\iff \cos 2x=\pm\frac{\sqrt 2}2.$$
Now

*

*$\cos 2x=\frac{\sqrt 2}2\iff 2x\equiv \pm\frac\pi 4\pmod{2\pi}\iff x\equiv \pm\frac\pi 8\pmod{\pi}$

*$\cos 2x=-\frac{\sqrt 2}2\iff 2x\equiv \pm\frac{3\pi}4\pmod{2\pi}\iff x\equiv \pm\frac{3\pi}8\pmod{\pi}$.

