Question Number 8273 by lepan last updated on 05/Oct/16
$${Express}\:{sin}\alpha+\sqrt{\mathrm{3}}{cos}\alpha\:{in}\:{the}\:{form}\: \\ $$ $${Rsin}\left(\alpha+\beta\right)\:{where}\:{R}>\mathrm{0}\:{and}\:\mathrm{0}°<\beta<\mathrm{90}°. \\ $$ $${Hence}\:{solve}\:{the}\:{equation}\:{sin}\alpha+\sqrt{\mathrm{3}}{cos}\alpha=\mathrm{2} \\ $$ $${for}\:\mathrm{0}°<\alpha<\mathrm{270}°. \\ $$
Answered by prakash jain last updated on 05/Oct/16
$$\mathrm{sin}\:\alpha+\sqrt{\mathrm{3}}\mathrm{cos}\:\alpha \\ $$ $$=\mathrm{2}\left(\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin}\:\alpha+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{cos}\:\alpha\right) \\ $$ $$=\mathrm{2}\left(\mathrm{sin}\:\frac{\pi}{\mathrm{6}}\mathrm{sin}\:\alpha+\mathrm{cos}\:\frac{\pi}{\mathrm{6}}\mathrm{cos}\:\alpha\right) \\ $$ $$=\mathrm{2cos}\:\left(\frac{\pi}{\mathrm{6}}−\alpha\right) \\ $$ $$\mathrm{sin}\:\alpha+\sqrt{\mathrm{3}}\mathrm{cos}\:\alpha=\mathrm{1} \\ $$ $$\Rightarrow\mathrm{2cos}\:\left(\frac{\pi}{\mathrm{6}}−\alpha\right)=\mathrm{1} \\ $$ $$\Rightarrow\mathrm{cos}\:\left(\frac{\pi}{\mathrm{6}}−\alpha\right)=\frac{\mathrm{1}}{\mathrm{2}} \\ $$ $$\Rightarrow\mathrm{cos}\:\left(\frac{\pi}{\mathrm{6}}−\alpha\right)=\mathrm{cos}\:\frac{\pi}{\mathrm{3}} \\ $$ $$\frac{\pi}{\mathrm{6}}−\alpha=\mathrm{2}{n}\pi\pm\frac{\pi}{\mathrm{3}} \\ $$ $$\alpha=−\mathrm{2}{n}\pi+\left(\frac{\pi}{\mathrm{6}}\pm\frac{\pi}{\mathrm{3}}\right)=\begin{cases}{\mathrm{2}{k}\pi+\frac{\pi}{\mathrm{2}}}\\{\mathrm{2}{k}\pi−\frac{\pi}{\mathrm{6}}}\end{cases} \\ $$ $$\mathrm{between}\:\mathrm{0}\:\mathrm{to}\:\mathrm{270}°\:\left({or}\:\mathrm{0}\:\mathrm{to}\:\frac{\mathrm{3}\pi}{\mathrm{2}}\right) \\ $$ $$\alpha=\frac{\pi}{\mathrm{2}}\mathrm{or}\:\mathrm{90}° \\ $$