Tricky trig problem If $\sin x \cos x + \sin y \cos y + \sin x \sin y + \cos x \cos y = 1$ and $\cos (x-y)$ is the smallest possible, what is the value of $2x-y$, expressed in degrees, that is closest 360º? (AMC 2012, Senior)
I tried using a number of trigonometric identities, going backwards from $\cos (x-y)$, but didn't really get anywhere. Other than that, I'm not too sure of how to approach this question.

The equation $\sin (x) \sin (y)+\cos (x) \cos (y)+\sin (x) \cos (x)+\sin (y) \cos (y)=1$ can be written in this way look here
$\cos(x-y) (\sin (x+y)+1)=1$
As they say that $\cos(x-y)$ is the minimum, then $\sin (x+y)+1$ is the maximum, which means that $\sin (x+y)=1$ and $\cos(x-y)=\dfrac{1}{2}$
Then we solve the systems
$$\left\{x-y=\frac{\pi }{3}+2 \pi  k,x+y=\pi  h+\frac{\pi }{2}\right\}$$
and
$$\left\{x-y=-\frac{\pi }{3}+2 \pi  k,x+y=\pi  h+\frac{\pi }{2}\right\}$$
which give infinitely many solutions
$$x= \frac{1}{12} (6 \pi  h+12 \pi  k+5 \pi ),y= \frac{1}{12} (6 \pi  h-12 \pi  k+\pi )$$
and
$$x\to \frac{1}{12} (6 \pi  h+12 \pi  k+\pi ),y\to \frac{1}{12} (6 \pi  h-12 \pi  k+5 \pi );\;k,\;h\in\mathbb{Z}$$
The value of $2x-y$ closest to 360° is $\dfrac{7\pi}{4}=315^{\circ}$
