Question

A value of x satisfying \[\cos x + \sqrt{3} \sin x = 2\] is

 

Options

• `(5pi)/3`

• \[\frac{4\pi}{3}\]

• `(2pi)/3`

• \[\frac{\pi}{3}\]

Answer 1

\[\frac{\pi}{3}\]
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 `sqrt3=r sinalpha`

\[\Rightarrow r = \sqrt{a^2 + b^2} = \sqrt{(\sqrt{3} )^2 + 1^2} = 2\] 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:
\[r \cos \alpha \cos x + r \sin \alpha \sin x = 2\]
\[ \Rightarrow r \cos\left( x - \alpha \right) = 2\]
\[ \Rightarrow r \cos\left( x - \frac{\pi}{3} \right) = 2\]
\[ \Rightarrow 2 \cos \left( x - \frac{\pi}{3} \right) = 2\]
\[ \Rightarrow \cos \left( x - \frac{\pi}{3} \right) = 1\]
\[ \Rightarrow \cos \left( x - \frac{\pi}{3} \right) = \cos 0\]
\[ \Rightarrow x - \frac{\pi}{3} = 0\]
\[ \Rightarrow x = \frac{\pi}{3}\]