hello-help-me-to-solve-it-in-please-x-y-6-cosx-3-cosy-1-2-

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

$$\left.\mathrm{h}\left.\mathrm{e}\left.\mathrm{l}\left.\mathrm{l}\left.\mathrm{o}\right]\:\mathrm{help}\:\mathrm{me}\:\mathrm{to}\:\mathrm{solve}\:\mathrm{it}\:\mathrm{in}\:\right]−\Pi;\Pi\right]×\right]−\Pi;\Pi\right]\:\mathrm{please} \\ $$$$\begin{cases}{\mathrm{x}−\mathrm{y}=\frac{\Pi}{\mathrm{6}}}\\{\mathrm{cosx}−\sqrt{\mathrm{3}}\mathrm{cosy}=−\frac{\mathrm{1}}{\mathrm{2}}}\end{cases} \\ $$

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

$${cos}\left(\frac{\pi}{\mathrm{6}}+{y}\right)−\sqrt{\mathrm{3}}\:{cosy}=\frac{−\mathrm{1}}{\mathrm{2}} \\ $$$$\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{cosy}−\frac{\mathrm{1}}{\mathrm{2}}{siny}−\sqrt{\mathrm{3}}\:{cosy}=\frac{−\mathrm{1}}{\mathrm{2}} \\ $$$$\frac{−\mathrm{1}}{\mathrm{2}}{siny}−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{cosy}=\frac{−\mathrm{1}}{\mathrm{2}} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}{siny}+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{cosy}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${sin}\left(\frac{\pi}{\mathrm{3}}+{y}\right)={sin}\frac{\pi}{\mathrm{6}} \\ $$$$\frac{\pi}{\mathrm{3}}+{y}=\frac{\pi}{\mathrm{6}}\rightarrow{y}=−\frac{\pi}{\mathrm{6}} \\ $$$${so}\:{x}={y}+\frac{\pi}{\mathrm{6}}\rightarrow{x}=\mathrm{0} \\ $$$${sin}\left(\frac{\pi}{\mathrm{3}}+{y}\right)={sin}\frac{\pi}{\mathrm{6}}={sin}\left(\pi−\frac{\pi}{\mathrm{6}}\right) \\ $$$$\frac{\pi}{\mathrm{3}}+{y}=\frac{\mathrm{5}\pi}{\mathrm{6}}\rightarrow{y}=\frac{\mathrm{5}\pi}{\mathrm{6}}−\frac{\mathrm{2}\pi}{\mathrm{6}}=\frac{\pi}{\mathrm{2}} \\ $$$${x}=\frac{\pi}{\mathrm{6}}+{y}\rightarrow{x}=\frac{\pi}{\mathrm{6}}+\frac{\mathrm{3}\pi}{\mathrm{6}}=\frac{\mathrm{4}\pi}{\mathrm{6}} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

Terms of Service
Privacy Policy
Contact: info@tinkutara.com