Question

The general value of x satisfying the equation
\[\sqrt{3} \sin x + \cos x = \sqrt{3}\]

विकल्प

• \[x = n\pi + \left( - 1 \right)^n \frac{\pi}{4} + \frac{\pi}{3}, n \in Z\]

   

• \[x = n\pi + \left( - 1 \right)^n \frac{\pi}{3} + \frac{\pi}{6}, n \in Z\]

• \[x = n\pi \pm \frac{\pi}{6}, n \in Z\]

   

• \[x = n\pi \pm \frac{\pi}{3}, n \in Z\]

Answer 1

\[x = n\pi + \left( - 1 \right)^n \frac{\pi}{3} - \frac{\pi}{6}, n \in Z\]
Given: 

\[\sqrt{3} \sin x + \cos x = \sqrt{3}\] ...(i)
This equation is of the form 

\[a \sin\theta + b \cos\theta = c\], where

\[a = \sqrt{3}, b = 1\] and \[c = \sqrt{3}\].
Let: a = r cos α and b = r sin α
Now,

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

\[\tan\alpha = \frac{b}{a} \Rightarrow \tan\alpha = \frac{1}{\sqrt{3}}\]

`=>alpha = pi/6` On putting \[a = \sqrt{3} = r \cos\alpha\] and \[b = 1 = r \sin\alpha\] in equation (i),  we get:
\[r \cos\alpha \sin x + r \sin\alpha \cos x = \sqrt{3}\]
\[ \Rightarrow r \sin (x + \alpha) = \sqrt{3}\]
\[ \Rightarrow 2 \sin ( x + \alpha) = \sqrt{3}\]
\[ \Rightarrow \sin (x + \alpha) = \frac{\sqrt{3}}{2}\]
\[ \Rightarrow \sin (x + \alpha) = \sin \frac{\pi}{3}\]
\[ \Rightarrow \sin \left( x + \frac{\pi}{6} \right) = \sin \frac{\pi}{3}\]
\[ \Rightarrow x = n\pi + ( - 1 )^n \frac{\pi}{3} - \frac{\pi}{6} , n \in Z\]