Question

If $\sin \theta +\sqrt{3}\cos \theta =\sqrt{2}$, then $\theta =\frac{5\pi }{m}$. Find $m$

Answer 1

Solution:-

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

$\sqrt{3}\cos \theta =\sqrt{2}-\sin \theta$

Squaring both sides, we have

${\left(\sqrt{3}\cos \theta \right)}^2={(\sqrt{2}-\sin \theta )}^2$

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

$\Rightarrow 3(1-{\sin }^2\theta )=2+{\sin }^2\theta -2\sqrt{2}\sin \theta (\because {\sin }^2\theta +{\cos }^2\theta =1)$

$3-3{\sin }^2\theta =2+{\sin }^2\theta -2\sqrt{2}\sin \theta$

$\Rightarrow 4{\sin }^2\theta -2\sqrt{2}\sin \theta -1=0$

The above equation is in the quadratic form, i.e. $a{x}^2+bx+c$.

Now, by using quadratic formula, i.e., $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$, we have

$\sin \theta =\frac{-(-2\sqrt{2})\pm \sqrt{{(-2\sqrt{2})}^2-(4\times 4\times (-1))}}{2\times 4}$

$\Rightarrow \sin \theta =\frac{2\sqrt{2}\pm \sqrt{8+16}}{8}$

$\Rightarrow \sin \theta =\frac{\sqrt{2}\pm \sqrt{6}}{4}$

$\therefore \theta ={\sin }^{-1}\left(\frac{\sqrt{2}+\sqrt{6}}{4}\right)+2n\pi \text{Or}\theta ={\sin }^{-1}\left(\frac{\sqrt{2}-\sqrt{6}}{4}\right)+2n\pi \text{For all n}\in \text{integer}$

If $\theta =\frac{5\pi }{m}$

Case I:-

${\sin }^{-1}\left(\frac{\sqrt{2}+\sqrt{6}}{4}\right)=\frac{5\pi }{m}$

$\Rightarrow m=\frac{5\pi }{{\sin }^{-1}\left(\frac{\sqrt{2}+\sqrt{6}}{4}\right)}$

Case II:-

${\sin }^{-1}\left(\frac{\sqrt{2}-\sqrt{6}}{4}\right)=\frac{5\pi }{m}$

$\Rightarrow m=\frac{5\pi }{{\sin }^{-1}\left(\frac{\sqrt{2}-\sqrt{6}}{4}\right)}$