# How to solve this system of two trigonometric equations

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.

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

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.

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