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Question Number 8273 by lepan last updated on 05/Oct/16

$$Expresssin\alpha +\sqrt{3}cos\alpha intheform$$ $$Rsin(\alpha +\beta )whereR>0and0°<\beta <90°.$$ $$Hencesolvetheequationsin\alpha +\sqrt{3}cos\alpha =2$$ $$for0°<\alpha <270°.$$

Answered by prakash jain last updated on 05/Oct/16

$$\sin \alpha +\sqrt{3}\cos \alpha$$ $$=2(\frac{1}{2}\sin \alpha +\frac{\sqrt{3}}{2}\cos \alpha )$$ $$=2(\sin \frac{\pi }{6}\sin \alpha +\cos \frac{\pi }{6}\cos \alpha )$$ $$=2\cos (\frac{\pi }{6}-\alpha )$$ $$\sin \alpha +\sqrt{3}\cos \alpha =1$$ $$\Rightarrow 2\cos (\frac{\pi }{6}-\alpha )=1$$ $$\Rightarrow \cos (\frac{\pi }{6}-\alpha )=\frac{1}{2}$$ $$\Rightarrow \cos (\frac{\pi }{6}-\alpha )=\cos \frac{\pi }{3}$$ $$\frac{\pi }{6}-\alpha =2n\pi \pm \frac{\pi }{3}$$ $$\alpha =-2n\pi +(\frac{\pi }{6}\pm \frac{\pi }{3})=\{\begin{array}{l}2k\pi +\frac{\pi }{2}\\ 2k\pi -\frac{\pi }{6}\end{array}$$ $$\mathrm{between}0\mathrm{to}270°\left(or0\mathrm{to}\frac{3\pi }{2}\right)$$ $$\alpha =\frac{\pi }{2}\mathrm{or}90°$$

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