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$x^2+y^2=12^2$ is a circle having centre at $(0,0)$ & radius $12$. So,I could easily mark the first inequality.

But $\sin(2x+3y)=0$ implies $2x+3y=n(\pi)$ where $n$ is any integer For $n=0$, $2x+3y=0$ & the graph is a straight line passing through origin and is easy to plot. But for other values of $n$, I'm unable to plot the graph and thus being unable to find the area.

Please help.

Thank You.

Gregory Grant
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Number
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  • Suppose you've plot $y=-\frac{2}{3}x$. Then $y=-\frac{2}{3}x+\frac{2}{3}\pi n$ is a line $y=-\frac{2}{3}x$ lifted up by $\frac{2}{3}\pi n$ units. http://www.wolframalpha.com/input/?i=plot+x%5E2%2By%5E2%E2%89%A4144+%26%26+sin%282x%2B3y%29%E2%89%A40 – Alexey Burdin May 14 '15 at 20:24
  • @Alexey Burdin:Yes.No problem when n=0,but when n=any other value say 2 then how to mark (4/3)pi without a graphing calculator,& how to find the area at the end.can you please explain. – Number May 14 '15 at 20:30
  • Why do so many people hate to put spaces after their punctuation? I see this dozens of times a day - not just as an isolated typo once or twice in a post, but many users do this as a rule. – Gregory Grant May 14 '15 at 20:34
  • @Number: the values of $\pi, e, \ln 2$,etc usually are plotted aproximately. Nevertheless, plot doesn't look nice, see the link above. The area is $1/2$ area of the disc i.e. $\frac{12^2\pi}{2}$. Guess why. :) – Alexey Burdin May 14 '15 at 20:36
  • @GregoryGrant one can start a meta post on this topic. If it were in the top, people would mind these spaces. :)) – Alexey Burdin May 14 '15 at 20:38
  • @AlexeyBurdin Thank you, not sure how to do that tho... I'll investigate – Gregory Grant May 14 '15 at 20:39

1 Answers1

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For every point $X(x,y)$ inside the circle we specify another point $Y(-x,-y)$. Such, that $\sin(2x+3y)\le 0$ for $X$, $\sin(-2x-3y)\ge 0$ for $Y$ and vice versa.
$f(X)=Y$ is one-to-one except $(0,0)$, so all the points are divided in two sets with equal area (indeed, applying $f$ is flipping across $x+y=0$) with intersection area $0$, so each occupy $1/2$ of the area of the disc.

Alexey Burdin
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