Question

If \[\sqrt{3} \cos x + \sin x = \sqrt{2}\] , then general value of ​x is

पर्याय

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

   

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

• \[n \pi + \frac{\pi}{4} - \frac{\pi}{3}, n \in Z\]

   

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

Answer 1

\[n \pi + \left( - 1 \right)^n \frac{\pi}{4} - \frac{\pi}{3}, n \in Z\]
Given equation:
\[\sqrt{3}\cos x + \sin x = \sqrt{2}\]  ...(i)
This is of the form \[a \cos x + b \sin x = c\], where
\[a = \sqrt{3} , b = 1\] and \[c = \sqrt{2}\].
Let: a = r sin α and b = r sin α.
Now,
\[r = \sqrt{a^2 + b^2} = \sqrt{(\sqrt{3} )^2 + 1^2} = 2\]
And,
\[\tan \alpha = \frac{a}{b} \]
\[ \Rightarrow \tan \alpha = \frac{\sqrt{3}}{1} \]
\[ \Rightarrow \tan \alpha = \tan \frac{\pi}{3} \]
\[ \Rightarrow \alpha = \frac{\pi}{3}\]
Putting 
\[a = \sqrt{3} = r \sin \alpha\] and \[b = 1 = r \cos \alpha\] in equation (i), we get:

\[r \cos x \sin\alpha + r \sin x \cos\alpha = \sqrt{2}\]

\[ \Rightarrow r \sin (x + \alpha) = \sqrt{2}\]

\[ \Rightarrow 2 \sin (x + \alpha) = \sqrt{2}\]

\[ \Rightarrow \sin \left( x + \frac{\pi}{3} \right) = \frac{1}{\sqrt{2}}\]

\[ \Rightarrow \sin \left( x + \frac{\pi}{3} \right) = \cos \frac{\pi}{4}\]

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

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