Вопрос в картинках...

Решите задачу:

$\left \{ {{ \frac{sinx}{siny} =1} \atop {x-y= \frac{\pi}{3 }} \right. \\\ \left \{ {{ sinx=siny} \atop {x=y+ \frac{\pi}{3 }} \right. \\\ sin(y+ \frac{\pi}{3 })=siny \\\ sinycos\frac{\pi}{3 }+cosysin\frac{\pi}{3 }=siny \\\ \frac{1}{2} siny+ \frac{ \sqrt{3} }{2}cosy=siny \\\ \frac{1}{2} siny- \frac{ \sqrt{3} }{2}cosy=0 \\\ siny- \sqrt{3} cosy=0 \\\ tg y-\sqrt{3}=0 \\\ y= \frac{\pi}{3}+\pi n , n\in Z \\\ x= \frac{\pi}{3}+ \frac{\pi}{3}+\pi k= \frac{2\pi}{3}+\pi k, k\in Z$