Level curve of $f(x,y) = cos(x+y)$

Question

Find and sketch the level curve of the function $f(x,y) = cos(x+y)$ when $f(x,y) = -1$

I've tried to see how this curve is using implicit derivative, but I have no clue of how this curve is.

Thank you very much!

Answer 1

If the sought-for curve is $\cos(x + y) = -1$, as in the title, then the answer is most easily had: $\cos(x + y) = -1$ implies $x + y = (2n + 1)\pi$, for any and all integers $n$. Thus the level curves are the lines $y = -x + (2n + 1)\pi$, which are the straight lines of slope $-1$ which intersect the $x$-axis at the points $(2n + 1)\pi, 0)$. They are infinite in number and all parallel to one another.

Hope this helps. Cheerio,

and as always,

Fiat Lux!!!

Answer 2

$\cos u=-1$ iff $u=\rm what?$. Then $\{(x,y):\cos(x+y)=-1\}=\{(x,y):x+y=\rm what?\}$