A value of x satisfying cos x

Solution:

(d) $\frac{\pi}{3}$

Given equation: 

$\cos x+\sqrt{3} \sin x=2 \quad \ldots$ (i)

Thus, the equation is of the form $a \cos x+b \sin x=c$, where $a=1, b=\sqrt{3}$ and $c=3$.

Let:

$a=r \cos \alpha$ and $b=r \sin \alpha$

$1=r \cos \alpha$ and $\sqrt{3}=r \sin \alpha$

$\Rightarrow r=\sqrt{a^{2}+b^{2}}=\sqrt{(\sqrt{3})^{2}+1^{2}}=2$ and

$\tan \alpha=\frac{b}{a} \Rightarrow \tan \alpha=\frac{\sqrt{3}}{1} \Rightarrow \tan \alpha=\tan \frac{\pi}{3} \Rightarrow \alpha=\frac{\pi}{3}$

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

$r \cos \alpha \cos x+r \sin \alpha \sin x=2$

$\Rightarrow r \cos (x-\alpha)=2$

$\Rightarrow r \cos \left(x-\frac{\pi}{3}\right)=2$

$\Rightarrow 2 \cos \left(x-\frac{\pi}{3}\right)=2$

$\Rightarrow \cos \left(x-\frac{\pi}{3}\right)=1$

$\Rightarrow \cos \left(x-\frac{\pi}{3}\right)=\cos 0$

$\Rightarrow x-\frac{\pi}{3}=0$

$\Rightarrow x=\frac{\pi}{3}$