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If $x + y = 3 - cos4\theta$ and $x - y = 4 \,sin2\theta$ then
$x^4 + y^4 = 9$
$\sqrt x \, + \,\sqrt y \, = \,16\,$
$x^3 + y^3 = 2(x^2 + y^2)$
$\sqrt x \, + \,\sqrt y \, = \,2$
Solution
On adding and subtracting
$x = \frac{{3 – \cos 4\theta \, + 4\sin 2\theta }}{2}\,$; $y = \frac{{3 – \cos 4\theta \, – 4\sin 2\theta }}{2}\,$
$x =\frac{{4(1 + \sin 2\theta )\, – \,(1 + \cos 4\theta )}}{2}\,$ ; $y =\frac{{4(1 – \sin 2\theta )\, – \,(1 + \cos 4\theta )}}{2}\,$
$x = 2 (1 + sin2\theta ) – cos^22\theta$ ; $y = 2 (1 – sin2\theta ) – cos^22\theta$
$x = 1 + 2\, sin2\theta + sin^22\theta$ ;$ y = 1 – 2 sin2\theta + sin^22\theta$
$x = (1 + sin2\theta )^2$ ;$y = (1 – sin2\theta )^2$ $\Rightarrow \sqrt x \, + \,\sqrt y \, = \,2$
Alternate : Or put $\theta$ = $\frac{\pi }{4}\,$ and verify