If cos x
Question:

If $\cos x+\sqrt{3} \sin x=2$, then $x=$

(a) $\pi / 3$

(b) $2 \pi / 3$

(c) $4 \pi / 6$

(d) $5 \pi / 12$

Solution:

(a) $\pi / 3$

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

This equation is of the form $a \cos x+b \sin x=c$, where $a=1, b=\sqrt{3}$ and $c=2$.

Let:

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

Now,

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

$\Rightarrow r=\sqrt{a^{2}+b^{2}}=\sqrt{1+3}=\sqrt{4}=2$

And,

$\tan \alpha=\frac{b}{a}$

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

$\Rightarrow \tan \alpha=\sqrt{3}$

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

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

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

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

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

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

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

$\Rightarrow x-\frac{\pi}{3}=2 n \pi \pm 0$

$\Rightarrow x=2 \mathrm{n} \pi \pm \frac{\pi}{3}$

For $n=0, x=\frac{\pi}{3}$.

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