Find all the tangents to the curve $y = \cos(x + y),x$ which belongs to $[-\pi$,$\pi]$, that are parallel to the line $x + 2y = 0$.
I differentiate it and equate to $(-1/2)$.
$$y' = -\sin(x+y)[1+y']$$
And I get $[\sin(x+y)]/[1+\sin(x+y)]= (1/2)$.
But then how to find the tangent?
Yes you can proceed continuously after that step. We have got the slope of tangent at $(x,y) = \frac{dy}{dx} = \frac{-\sin(x+y)}{ 1+\sin(x+y)}$.
Since tangents to the given curve are parallel to $x +2y =0$ which has a slope of $-\frac{1}{2}$, we can get $$\sin(x +y) =1$$ $$\Rightarrow x+y = n\pi +(-1)^n\frac{\pi}{2} , n\in \mathbb Z$$ Then $y =0$ for all $n \in \mathbb Z$.
Since, $-\pi \leq x \leq \pi$, we get $x =\frac{\pi}{2}$. Thus, tangents to the given curve are parallel to $x+2y =0$ only at $(\frac{\pi}{2}, 0)$ for $-\pi \leq x \leq \pi$. Now, I leave it to you to find the tangent at $(\frac{\pi}{2}, 0)$. Hope it helps.