How to solve this system of two trigonometric equations

Question

I was trying to solve problem I found in one book, but I got into step from where I couldn't continue working on.

Here is what I got so far:

$$\tan(\pi/4+x) = 2\sqrt{2}\cos^3y \\ \cot(\pi/4+x) = 2\sqrt{2}\sin^3y $$

If we multiply the two equations we get

$1 = 8\sin^3y\cos^3y\\1=2\sin y\cos y$

And from here I don't know which formulas should be used to solve for y, please give me some hints on how to continue solving the system.

Answer 1

$2\sin{y}\cos{y}=\sin{2y}=1$ and therefore $2y=\frac{\pi}{2}\bmod {2\pi}$ and $y=\frac{\pi}{4}\bmod {\pi}$

Answer 2

Hint:

$$\implies\sin2y=1\implies2y=2n\pi+\dfrac\pi2$$ where $n$ is any integer

$$\implies\cos y=\cos\left(n\pi+\dfrac\pi4\right)=(-1)^n\dfrac1{\sqrt2}$$

$$\tan\left(\dfrac\pi4+x\right)=(-1)^n$$

Check for even$=(2m)$/odd$(=2m+1)$ values of $n$ separately.

Answer 3

Hint: $$\tan(\frac{\pi}{4}+x)=\frac{1+\tan(x)}{1-\tan(x)}$$ and $$\cot(\frac{\pi}{4}+x)=\frac{\cot(x)-1}{1+\cot(x)}$$