Trigonometry! #131

The equation can be rewritten as

$\sin(x) + \sqrt{3}\cos(x) = -\sin(y) + \sqrt{3}\cos(y)$

$\Longrightarrow \dfrac{1}{2}\sin(x) + \dfrac{\sqrt{3}}{2}\cos(x) = -\dfrac{1}{2}\sin(y) + \dfrac{\sqrt{3}}{2}\cos(y)$

$\Longrightarrow \cos(\dfrac{\pi}{3})\sin(x) + \sin(\dfrac{\pi}{3})\cos(x) = \cos(\dfrac{2\pi}{3})\sin(y) + \sin(\dfrac{2\pi}{3})\cos(y)$

$\Longrightarrow \sin(x + \dfrac{\pi}{3}) = \sin(y + \dfrac{2\pi}{3})$ ,

and so either

(i) $x + \dfrac{\pi}{3} = y + \dfrac{2\pi}{3} + 2n\pi \Longrightarrow x = y + \dfrac{\pi}{3} + 2n\pi$ for any integer $n,$ or

(ii) $x + \dfrac{\pi}{3} = \pi - (y + \dfrac{2\pi}{3}) + 2n\pi \Longrightarrow x = -y + 2n\pi$ for any integer $n$ .

In case (i) we would then have $\sin(3x) = \sin(3y + \pi + 6n\pi) = -\sin(3y)$ , and in case (ii) we would then have $\sin(3x) = \sin(-3y + 6n\pi) = -\sin(3y)$ , and thus in either case we find that

$\sin(3y) + \sin(3x) = \sin(3y) - \sin(3y) = \boxed{0}.$

A Solution That Got Lucky

An Alternative Solution To $x=-y$

Edit: A little correction to the second image. The solution is true if $x=-y$ .

Just by looking the expression, if angles x=y=0 or n(pi) then cos(x)=cos(y)=1 and since cos(x) is subtracted from cos(y) the equation balance 0=0