Inspired by MIT Problem (III)

Relevant wiki: Half Angle Tangent Substitution

$\begin{cases} x + \sin y = a & ...(1) \\ x + a\cos y = a-1 &...(2) \end{cases} \quad \small \color{#3D99F6} \text{where }a = 2018$

$\begin{aligned} (1) - (2): \quad \sin y - a \cos y & = 1 & \small \color{#3D99F6} \text{Let }t = \tan \frac y2 \\ \frac {2t}{1+t^2} - \frac {a(1-t^2)}{1+t^2} & = 1 & \small \color{#3D99F6} \text{Multiply both sides by }1+t^2 \\ 2t - a + at^2 & = 1 + t^2 \\ (a-1)t^2 + 2t - a-1 & = 0 \\ ((a-1)t + a+1)(t-1) & = 0 \\ t & = \tan \frac y2 = 1 & \small \color{#3D99F6} \text{For }0 \le y \le \frac \pi 2, \ \tan \frac y2 > 0 \\ \frac y 2 & = \frac \pi 4 & \small \color{#3D99F6} \text{Putting }y = \frac \pi 2 \text{ in (2).} \\ x + 0 & = a - 1 = 2017 \end{aligned}$

Therefore, $x+y = \boxed{2017+ \dfrac \pi 2}$ .

Notice that all the answers indicate that $y=\frac{\pi}{2}\implies \cos y=0 \text{ and } \sin y= 1$ , plugging those values in the two equations and solving for $x \implies x=2017.$

Another Solution: subtracting the two equations will give $(\sin y-2018\cos y= 1)$ , since $0\le y \le \frac{\pi}{2}$ ; therefore, the maximum of $y$ is 1 and the minimum is $0 \implies y=\frac{\pi}{2} \implies x=2017.$