Question

The general value of x satisfying the equation $\sqrt{3}\sin x+\cos x=\sqrt{3}$ 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) $x=n\pi \pm \frac{\pi }{6},n\in Z$

(d) $x=n\pi \pm \frac{\pi }{3},n\in Z$

Answer 1

(b) $x=n\pi +{\left(-1\right)}^n\frac{\pi }{3}-\frac{\pi }{6},n\in Z$

Given:
$\sqrt{3}\sin x+\cos x=\sqrt{3}$ ...(i)
This equation is of the form $a\sin \theta +b\cos \theta =c$, where $a=\sqrt{3},b=1$ and $c=\sqrt{3}$.
Let:
$a=r\cos \alpha$ and $b=r\sin \alpha$
Now,
$r=\sqrt{a^2+b^2}=\sqrt{(\sqrt{3}{)}^2+1^2}=2$ and $\tan \alpha =\frac{b}{a}\Rightarrow \tan \alpha =\frac{1}{\sqrt{3}}$ $\Rightarrow \alpha =\frac{\mathrm{\pi }}{6}$
On putting $a=\sqrt{3}=r\cos \alpha$ and $b=1=r\sin \alpha$ in equation (i), we get:

$$\begin{gathered} r\cos \alpha \sin x+r\sin \alpha \cos x=\sqrt{3} \\ \Rightarrow r\sin (x+\alpha )=\sqrt{3} \\ \Rightarrow 2\sin (x+\alpha )=\sqrt{3} \\ \Rightarrow \sin (x+\alpha )=\frac{\sqrt{3}}{2} \\ \Rightarrow \sin (x+\alpha )=\sin \frac{\mathrm{\pi }}{3} \\ \Rightarrow \sin \left(x+\frac{\mathrm{\pi }}{6}\right)=\sin \frac{\mathrm{\pi }}{3} \\ \Rightarrow x=n\mathrm{\pi }+(-1{)}^{\mathrm{n}}\frac{\mathrm{\pi }}{3}-\frac{\mathrm{\pi }}{6},\mathrm{n}\in \mathrm{Z} \end{gathered}$$