Question

A value of x satisfying $\cos x+\sqrt{3}\sin x=2$ is
(a) $\frac{5\pi }{3}$

(b) $\frac{4\pi }{3}$

(c) $\frac{2\pi }{3}$​

(d) $\frac{\pi }{3}$​

Answer 1

(d) $\frac{\pi }{3}$
Given equation:
$\cos x+\sqrt{3}\sin x=2$ ...(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 $\tan \alpha =\frac{b}{a}\Rightarrow \tan \alpha =\frac{\sqrt{3}}{1}\Rightarrow \tan \alpha =\tan \frac{\mathrm{\pi }}{3}\Rightarrow \alpha =\frac{\mathrm{\pi }}{3}$
On putting $a=1=r\cos \alpha$ and $b=\sqrt{3}=r\sin \alpha$ in equation (i), we get:

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