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Question Number 74130 by mathocean1 last updated on 19/Nov/19

$$\mathrm{h}\mathrm{e}\mathrm{l}\mathrm{l}\mathrm{o}]\mathrm{help}\mathrm{me}\mathrm{to}\mathrm{solve}\mathrm{it}\mathrm{in}]-\mathrm{\Pi };\mathrm{\Pi }]\times ]-\mathrm{\Pi };\mathrm{\Pi }]\mathrm{please}$$$$\{\begin{array}{l}\mathrm{x}-\mathrm{y}=\frac{\mathrm{\Pi }}{6}\\ \mathrm{cosx}-\sqrt{3}\mathrm{cosy}=-\frac{1}{2}\end{array}$$

Answered by Tanmay chaudhury last updated on 19/Nov/19

$$cos(\frac{\pi }{6}+y)-\sqrt{3}cosy=\frac{-1}{2}$$$$\frac{\sqrt{3}}{2}cosy-\frac{1}{2}siny-\sqrt{3}cosy=\frac{-1}{2}$$$$\frac{-1}{2}siny-\frac{\sqrt{3}}{2}cosy=\frac{-1}{2}$$$$\frac{1}{2}siny+\frac{\sqrt{3}}{2}cosy=\frac{1}{2}$$$$sin(\frac{\pi }{3}+y)=sin\frac{\pi }{6}$$$$\frac{\pi }{3}+y=\frac{\pi }{6}\to y=-\frac{\pi }{6}$$$$sox=y+\frac{\pi }{6}\to x=0$$$$sin(\frac{\pi }{3}+y)=sin\frac{\pi }{6}=sin(\pi -\frac{\pi }{6})$$$$\frac{\pi }{3}+y=\frac{5\pi }{6}\to y=\frac{5\pi }{6}-\frac{2\pi }{6}=\frac{\pi }{2}$$$$x=\frac{\pi }{6}+y\to x=\frac{\pi }{6}+\frac{3\pi }{6}=\frac{4\pi }{6}$$$$$$$$$$$$$$$$$$$$$$

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