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Question

# The general value of x satisfying the equation $\sqrt{3}\mathrm{sin}x+\mathrm{cos}x=\sqrt{3}$ is given by (a) $x=n\pi +{\left(-1\right)}^{n}\frac{\pi }{4}+\frac{\pi }{3},n\in Z$ (b) $x=n\pi +{\left(-1\right)}^{n}\frac{\pi }{3}+\frac{\pi }{6},n\in Z$ (c) $x=n\pi ±\frac{\pi }{6},n\in Z$ (d) $x=n\pi ±\frac{\pi }{3},n\in Z$

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Solution

## (b) $x=n\pi +{\left(-1\right)}^{n}\frac{\pi }{3}-\frac{\pi }{6},n\in Z$ Given: $\sqrt{3}\mathrm{sin}x+\mathrm{cos}x=\sqrt{3}$ ...(i) This equation is of the form $a\mathrm{sin}\theta +b\mathrm{cos}\theta =c$, where $a=\sqrt{3},b=1$ and $c=\sqrt{3}$. Let: $a=r\mathrm{cos}\alpha$ and $b=r\mathrm{sin}\alpha$ Now, $r=\sqrt{{a}^{2}+{b}^{2}}=\sqrt{\left(\sqrt{3}{\right)}^{2}+{1}^{2}}=2$ and $\mathrm{tan}\alpha =\frac{b}{a}⇒\mathrm{tan}\alpha =\frac{1}{\sqrt{3}}$ $⇒\alpha =\frac{\mathrm{\pi }}{6}$ On putting $a=\sqrt{3}=r\mathrm{cos}\alpha$ and $b=1=r\mathrm{sin}\alpha$ in equation (i), we get: $r\mathrm{cos}\alpha \mathrm{sin}x+r\mathrm{sin}\alpha \mathrm{cos}x=\sqrt{3}\phantom{\rule{0ex}{0ex}}⇒r\mathrm{sin}\left(x+\alpha \right)=\sqrt{3}\phantom{\rule{0ex}{0ex}}⇒2\mathrm{sin}\left(x+\alpha \right)=\sqrt{3}\phantom{\rule{0ex}{0ex}}⇒\mathrm{sin}\left(x+\alpha \right)=\frac{\sqrt{3}}{2}\phantom{\rule{0ex}{0ex}}⇒\mathrm{sin}\left(x+\alpha \right)=\mathrm{sin}\frac{\mathrm{\pi }}{3}\phantom{\rule{0ex}{0ex}}⇒\mathrm{sin}\left(x+\frac{\mathrm{\pi }}{6}\right)=\mathrm{sin}\frac{\mathrm{\pi }}{3}\phantom{\rule{0ex}{0ex}}⇒x=n\mathrm{\pi }+\left(-1{\right)}^{\mathrm{n}}\frac{\mathrm{\pi }}{3}-\frac{\mathrm{\pi }}{6},\mathrm{n}\in \mathrm{Z}$

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