The locus of points (x, y) in the plane satisfying sin2 x + sin2 y = 1 consists of
infinitely many lines with slope ± 1
The equation is a trigonometric equation in two variables; reducing it to a product of linear factors in cosines shows the locus is a union of straight lines.
Step 1 — Rewrite: \(\sin^{2}x = 1 - \sin^{2}y = \cos^{2}y\).
Step 2 — Use double angles: \(\dfrac{1-\cos 2x}{2} = \dfrac{1+\cos 2y}{2}\), so \(\cos 2x + \cos 2y = 0\), i.e. \(2\cos(x+y)\cos(x-y) = 0\).
Step 3 — Hence \(x + y = \dfrac{\pi}{2} + k\pi\) or \(x - y = \dfrac{\pi}{2} + k\pi\), for every integer k.
Step 4 — These are two families of parallel lines with slopes −1 and +1, one line for each integer k, so there are infinitely many of them.
The locus contains no circle (it is a union of straight lines), and the number of lines is not finite since k ranges over all integers.
infinitely many lines with slope ± 1 option is correct.
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