Question

If $\cos x+\sqrt{3}\sin x=2,\mathrm{then}x=$
(a) $\pi /3$
(b) $2\pi /3$
(c) $4\pi /6$
(d) $5\pi /12$

Answer 1

(a) $\pi /3$
Given: $\cos x+\sqrt{3}\sin x=2$ ...(i)
This equation is of the form $a\cos x+b\sin x=c$, where $a=1,b=\sqrt{3}$ and $c=2$.
Let:
$a=r\cos \alpha \mathrm{and}b=\sin \alpha$
Now,
$1=r\cos \alpha ,\sqrt{3}=r\sin \alpha$
$\Rightarrow r=\sqrt{a^2+b^2}=\sqrt{1+3}=\sqrt{4}=2$
And,
$$\begin{gathered} \tan \alpha =\frac{b}{a} \\ \Rightarrow \tan \alpha =\frac{\sqrt{3}}{1} \\ \Rightarrow \tan \alpha =\sqrt{3} \end{gathered}$$
$\Rightarrow \alpha =\frac{\mathrm{\pi }}{3}$
On putting $a=1=r\cos \alpha \mathrm{and}b=\sqrt{3}=r\sin \alpha$ in equation (i), we get:

$$\begin{gathered} r\cos x\cos \alpha +r\sin x\sin \alpha =2 \\ \Rightarrow r\cos (x-\alpha )=2 \\ \Rightarrow 2\cos \left(x-\frac{\mathrm{\pi }}{3}\right)=2 \\ \Rightarrow \cos \left(x-\frac{\mathrm{\pi }}{3}\right)=1 \\ \Rightarrow \cos \left(x-\frac{\mathrm{\pi }}{3}\right)=\cos 0 \\ \Rightarrow x-\frac{\mathrm{\pi }}{3}=2n\mathrm{\pi }\pm 0 \\ \Rightarrow x=2\mathrm{n\pi }\pm \frac{\mathrm{\pi }}{3} \end{gathered}$$

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