Question

If \[\cos x + \sqrt{3} \sin x = 2,\text{ then }x =\]

 

Options

• \[\pi/3\]

   

• \[2\pi/3\]

   

• \[4\pi/6\]

   

• \[5\pi/12\]

   

Answer 1

`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: \[a = r \cos \alpha\text{ and }b = \sin \alpha\]
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 \[a = 1 = r \cos \alpha\text{ and }b = \sqrt{3} = r \sin \alpha\] in equation (i), we get:

\[r \cos x \cos \alpha + r \sin x \sin \alpha = 2\]

\[ \Rightarrow r \cos ( x - \alpha) = 2\]

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

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

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

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

\[ \Rightarrow x = 2n\pi \pm \frac{\pi}{3}\]

For n = 0, x = `pi/3`

`therefore x= pi/3`