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# The general value of x satisfying the equation

Question:

The general value of $x$ satisfying the equation $\sqrt{3} \sin x+\cos x=\sqrt{3}$ is given by

(a) $x=n \pi+(-1)^{n} \frac{\pi}{4}+\frac{\pi}{3}, n \in Z$

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

(c) $x=n \pi \pm \frac{\pi}{6}, n \in Z$

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

Solution:

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

Given:

$\sqrt{3} \sin x+\cos x=\sqrt{3} \ldots$ (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 $\tan \alpha=\frac{b}{a} \Rightarrow \tan \alpha=\frac{1}{\sqrt{3}} \Rightarrow \alpha=\frac{\pi}{6}$

On putting $a=\sqrt{3}=r \cos \alpha$ and $b=1=r \sin \alpha$ in equation (i), we get:

$r \cos \alpha \sin x+r \sin \alpha \cos x=\sqrt{3}$

$\Rightarrow r \sin (x+\alpha)=\sqrt{3}$

$\Rightarrow 2 \sin (x+\alpha)=\sqrt{3}$

$\Rightarrow \sin (x+\alpha)=\frac{\sqrt{3}}{2}$

$\Rightarrow \sin (x+\alpha)=\sin \frac{\pi}{3}$

$\Rightarrow \sin \left(x+\frac{\pi}{6}\right)=\sin \frac{\pi}{3}$

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