All the points (x, y) in the plane satisfying | vedclass

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(A) Given the equation: $x^2+2x \sin(xy)+1=0$We can rewrite this as: $x^2+2x \sin(xy)+\sin^2(xy)+1-\sin^2(xy)=0$This simplifies to: $(x+\sin(xy))^2+\c

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a pair of straight lines

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(A) Given the equation: $x^2+2x \sin(xy)+1=0$We can rewrite this as: $x^2+2x \sin(xy)+\sin^2(xy)+1-\sin^2(xy)=0$This simplifies to: $(x+\sin(xy))^2+\cos^2(xy)=0$Since the sum of two squares is zero,each term must be zero:$(x+\sin(xy))^2=0 \Rightarrow x+\sin(xy)=0$$\cos^2(xy)=0 \Rightarrow \cos(xy)=0$From $\cos(xy)=0$,we have $xy = (2n+1)\frac{\pi}{2}$ for some integer $n$.Substituting $\sin(xy) = \pm 1$ into $x+\sin(xy)=0$,we get $x = \mp 1$.If $\sin(xy)=1$,then $x=-1$. Since $xy = (2n+1)\frac{\pi}{2}$,we get $y = -(2n+1)\frac{\pi}{2}$.If $\sin(xy)=-1$,then $x=1$. Since $xy = (2n+1)\frac{\pi}{2}$,we get $y = (2n+1)\frac{\pi}{2}$.These represent a set of parallel lines $x=1$ and $x=-1$,which are a pair of straight lines.

All the points $(x, y)$ in the plane satisfying the equation $x^2+2x \sin(xy)+1=0$ lie on

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