Question

Solve the following equation:

\[\sqrt{3} \cos x + \sin x = 1\]

Answer 1

 Given:

\[\sqrt{3} \cos x + \sin x = 1\] ...(i)
The equation is of the form of \[a \cos x + b \sin x = c\], where

\[a = \sqrt{3}, b = 1\] and C = 1.

Let: q = r cos α and \[a = r \cos \alpha\]

Now,

\[r = \sqrt{a^2 + b^2} = \sqrt{(\sqrt{3} )^2 + 1^2} = 2\] and

\[\tan \alpha = \frac{b}{a} = \frac{1}{\sqrt{3}} \Rightarrow \alpha = \frac{\pi}{6}\]
On putting

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

\[r \cos \alpha \cos x + r \sin \alpha \sin x = 1\]

\[\Rightarrow r \cos (x - \alpha) \hspace{0.167em} = 1\]

\[ \Rightarrow 2 \cos (x - \alpha) = 1\]

\[ \Rightarrow \cos \left( x - \frac{\pi}{6} \right) = \frac{1}{2}\]

\[ \Rightarrow \cos \left( x - \frac{\pi}{6} \right) = \cos \frac{\pi}{3}\]

\[ \Rightarrow x - \frac{\pi}{6} = 2n\pi \pm \frac{\pi}{3}, n \in Z\]

On taking positive sign, we get:

\[x - \frac{\pi}{6} = 2n\pi + \frac{\pi}{3} \]
\[ \Rightarrow x = 2n\pi + \frac{\pi}{3} + \frac{\pi}{6}\]
\[ \Rightarrow x = 2n\pi + \frac{\pi}{2}, n \in Z\]
\[ \Rightarrow x = (4n + 1)\frac{\pi}{2}, n \in Z\]

Now, on taking negative sign of the equation, we get:
\[x - \frac{\pi}{6} = 2m\pi - \frac{\pi}{3}, m \in Z\]
\[ \Rightarrow x = 2m\pi - \frac{\pi}{3} + \frac{\pi}{6}, m \in Z\]
\[ \Rightarrow x = 2m\pi - \frac{\pi}{6} = (12m - 1) \frac{\pi}{6}, m \in Z\]