Question

If $\sqrt{3}\cos x+\sin x=\sqrt{2}$, then general value of ​x is
(a) $n\pi +{\left(-1\right)}^n\frac{\pi }{4},n\in Z$

(b) ${\left(-1\right)}^n\frac{\pi }{4}-\frac{\pi }{3},n\in Z$

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

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

Answer 1

(d) $n\pi +{\left(-1\right)}^n\frac{\pi }{4}-\frac{\pi }{3},n\in Z$
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:
$a=r\sin \alpha$ and $b=r\cos \alpha$.
Now,
$r=\sqrt{a^2+b^2}=\sqrt{(\sqrt{3}{)}^2+1^2}=2$
And,
$$\begin{gathered} \tan \alpha =\frac{a}{b} \\ \Rightarrow \tan \alpha =\frac{\sqrt{3}}{1} \\ \Rightarrow \tan \alpha =\tan \frac{\mathrm{\pi }}{3} \\ \Rightarrow \alpha =\frac{\mathrm{\pi }}{3} \end{gathered}$$
Putting $a=\sqrt{3}=r\sin \alpha$ and $b=1=r\cos \alpha$ in equation (i), we get:
$$\begin{gathered} r\cos x\sin \alpha +r\sin x\cos \alpha =\sqrt{2} \\ \Rightarrow r\sin (x+\alpha )=\sqrt{2} \\ \Rightarrow 2\sin (x+\alpha )=\sqrt{2} \\ \Rightarrow \sin \left(x+\frac{\mathrm{\pi }}{3}\right)=\frac{1}{\sqrt{2}} \\ \Rightarrow \sin \left(x+\frac{\mathrm{\pi }}{3}\right)=\cos \frac{\mathrm{\pi }}{4} \\ \Rightarrow x+\frac{\mathrm{\pi }}{3}=n\mathrm{\pi }+(-1{)}^n\frac{\mathrm{\pi }}{4},\mathrm{n}\in \mathrm{Z} \\ \Rightarrow x=\mathrm{n\pi }+(-1{)}^n\frac{\mathrm{\pi }}{4}-\frac{\mathrm{\pi }}{3},\mathrm{n}\in \mathrm{Z} \end{gathered}$$