Trigonometry III

Let $x=rcos\theta$ and $y=rsin\theta$ for $r=\frac{1995\pi}{4}$ Parametrize the Solutions on the Circle Given by the Second Equation . Then $\phi=x+y=r\sqrt2sin(\theta + \frac{\pi}{4}$ Has Equal and cosine if $\phi \equiv \frac{\pi}{4}$ (mod $\pi$ ) , $\iff\frac{\phi-\frac\pi4}{\pi}\in\mathbb{Z}$ $\iff\phi=\phi_k=\frac\pi4(4k+1)$ for $k\in\mathbb{Z}$ .

But $\left|\frac{\phi}{r\sqrt2}\right|\le1\iff$ = $\left|4k+1\right|\le\frac{4r\sqrt2}{\pi}=1995\sqrt2\iff$ $\left|k+\frac14\right|\le\frac{r\sqrt2}{\pi}=\frac{1995\sqrt2}{4}\approx705.339$ , $\iff k\in\{-705,\dots,705\}$ , Which Entails $1411$ Distinct, Admisssible Values of $\phi$ .

To Each One of These Corresponds A Line of Slope $-1$ Which Intersects Our Circle at Two Points. So Each $\phi_k\in(-1,1)$ Furnishes Two Solutions $(x,y),(y,x)$ With $\theta=\theta_k=-\frac\pi4+\arcsin\frac{\phi_k}{r\sqrt2}$ and $(y,x)$ Representing $\theta=\frac\pi2-\theta_k$ , While if Either of $\phi=\pm1$ Were Admissible (Which Isn't the Case), It Would Only Add One New Solution. Thus We Have $2 \times 1411=2822$ Total Solutions.