= Solution
{c}
Use the <change of variables> $u=x+y$. Then $u'=1+y'=1+\cos u=2\cos^2(u/2)$, a <separable differential equation>. Where $u$ is not an odd multiple of $\pi$, <separation of variables> gives
$$
\int\frac{du}{1+\cos u}=\int dx,
\qquad
\boxed{\tan\frac{x+y}{2}=x+C.}
$$
Indeed the <derivative> of $\tan(u/2)$ with respect to $u$ is $1/(1+\cos u)$. The nonconstant branches can also be written $y=2\arctan(x+C)+2k\pi-x$, $k\in\mathbb Z$. Dividing by $1+\cos u$ excluded its zeros: the missing <equilibrium solutions> for $u$ give the additional <solutions of a differential equation>
$$
\boxed{y=(2k+1)\pi-x,\qquad k\in\mathbb Z.}
$$
They satisfy $y'=-1=\cos(x+y)$ directly. The transformed right side is continuously differentiable, so the <uniqueness theorem for ordinary differential equations> prevents a nonconstant branch from crossing an <equilibrium solution> at finite $x$.
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