Question

Solve $\sqrt{3}\cos \theta +\sin \theta =\sqrt{2}$

• A. $\theta =2n\pi +\frac{5\pi }{6}$ where $nϵZ$
• B. $\theta =2n\pi -\frac{\pi }{3}$, where $nϵZ$
• C. $\theta =2n\pi +\frac{5\pi }{12}$ where $nϵZ$
• D. $\theta =2n\pi -\frac{\pi }{12}$, where $nϵZ$

Answer 1

Answer: (C), (D)

The correct options are
C $\theta =2n\pi +\frac{5\pi }{12}$ where $nϵZ$
D $\theta =2n\pi -\frac{\pi }{12}$, where $nϵZ$

We have $\sqrt{3}\cos \theta +\sin \theta =\sqrt{2}$ (i)

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

Let $\sqrt{3}=r\cos \alpha$ and $1=r\sin \alpha$. Then

$r=\sqrt{a^2+b^2}=\sqrt{{\left(\sqrt{3}\right)}^2+1^2}=2$

and $\tan \alpha =\frac{1}{\sqrt{3}}$ or $\alpha =\frac{\pi }{6}$

Substituting
$\sqrt{3}=r\cos \alpha$ and $1=r\sin \alpha$ in Eq. (i), it
reduces to $r\cos \alpha \cos \theta +r\sin \theta =\sqrt{2}$. Thus,

$r\cos (\theta -\alpha )=\sqrt{2}$

$2\cos (\theta -\frac{\pi }{6})=\sqrt{2}$

$\cos (\theta -\frac{\pi }{6})=\frac{1}{\sqrt{2}}=\cos \frac{\pi }{4}$

$\Rightarrow$ $\theta -\frac{\pi }{6}=2n\pi \pm \frac{\pi }{4}$, $nϵZ$

$\theta =2n\pi \pm \frac{\pi }{4}+\frac{\pi }{6}$, $nϵZ$

$=2n\pi +\frac{\pi }{4}+\frac{\pi }{6}$ or

$\theta =2n\pi -\frac{\pi }{4}+\frac{\pi }{6}$

$=2n\pi +\frac{5\pi }{12}$ or $\theta =2n\pi -\frac{\pi }{12}$, where $nϵZ$