Simultaneous equations with a parameter

[mathlover2358]

Hello. I've been trying for some time but I cannot figure out how to solve this :/
The question is what value of a gives a unique solution?

sinx + siny = 0.25
cosx + cosy = a

 

[Romsek]

We're not supposed to just give you answers but I found this problem interesting so you lucked out.

[MATH]\sin(x) + \sin(y) = \dfrac 1 4\\ \sin^2(x) + \sin^2(y) + 2\sin(x)\sin(y) = \dfrac{1}{16}\\~\\ \cos(x) + \cos(y) = a\\ \cos^2(x) + \cos^2(y) + 2\cos(x)\cos(y) = a^2\\~\\ \text{now add them}\\ 1 + 1 + 2(\sin(x)\sin(y) + \cos(x)\cos(y) = a^2 + \dfrac{1}{16}\\ 2(1 + cos(x + y)) = a^2 + \dfrac{1}{16}\\ \cos(x+y) = \dfrac{a^2 + \frac{1}{16}}{2}-1\\ y = -x + \cos^{-1}\left(\dfrac{a^2 + \frac{1}{16}}{2}-1\right),~-x - \cos^{-1}\left(\dfrac{a^2 + \frac{1}{16}}{2}-1\right)\\ \text{In order for the solution to be unique we must have both arccosine terms evaluate to zero}\\ \text{This means that the argument of both must be 1}\\ \dfrac{a^2 + \frac{1}{16}}{2}-1 = 1\\ a = \pm \dfrac{3\sqrt{7}}{4} [/MATH]

 

[mathlover2358]

Thank You so much

 

[apple2357]

I wondering if there is a geometric way to tackle this. I will have a think!

Can't see an easy way but tried sketching them implicitly

 

[apple2357]

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