If

Solution:

(d) $n \pi+(-1)^{n} \frac{\pi}{4}-\frac{\pi}{3}, n \in Z$

Given equation: 

$\sqrt{3} \cos x+\sin x=\sqrt{2} \ldots$.(i)

This is of the form $a \cos x+b \sin x=c$, where $a=\sqrt{3}, b=1$ and $c=\sqrt{2}$.

Let:

$a=r \sin \alpha$ and $b=r \cos \alpha$.

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 $a \stackrel{3}{=} \sqrt{3}=r \sin \alpha$ and $b=1=r \cos \alpha$ in equation (i), we get:

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

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

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

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

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

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

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