Find the general solution of each of the following equations:

To Find: General solution.

Given: $\sqrt{3} \cos x+\sin x=1 \Rightarrow \cos \left(x-\frac{\pi}{6}\right)=\frac{1}{2}=\cos \left(\frac{\pi}{3}\right)$ or $\cos \left(\frac{5 \pi}{3}\right)$

[Divide $\sqrt{2}$ on both sides and $\cos (x-y)=\cos x \cos y-\sin x \sin y$ ]

Formula used: $\cos \theta=\cos \alpha \Rightarrow \theta=2 n \pi \pm \alpha$

By using above formula, we have

$\Rightarrow x-\frac{\pi}{6}=2 n \pi \pm \frac{\pi}{3} \Rightarrow x=2 n \pi \pm \frac{\pi}{3}+\frac{\pi}{6}$

$\Rightarrow x=2 n \pi+\frac{\pi}{2}$ or $x=2 n \pi-\frac{\pi}{6}$ where $n \in I$

So general solution is $x=2 n \pi+\frac{\pi}{2}$ or $x=2 n \pi-\frac{\pi}{6}$ where $n \in I$