Question

The general solution of $(\sqrt{3}-1)\sin \theta +(\sqrt{3}+1)\cos \theta =2$ $\Rightarrow \theta =2n\pi +\frac{\pi }{m}$ or $\theta =2n\pi -\frac{\pi }{n}$ Then n/m =

Answer 1

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

Divide the entire equation $2\sqrt{2},$ we get

$\Rightarrow \left(\frac{\sqrt{3}-1}{2\sqrt{2}}\right)\sin \theta +\left(\frac{\sqrt{3}+1}{2\sqrt{2}}\right)\cos \theta =\frac{1}{\sqrt{2}}$

$\Rightarrow \sin 15°\sin \theta +\cos 15°\cos \theta =\frac{1}{\sqrt{2}}$

$[\because \sin 15°=\frac{\sqrt{3}-1}{2\sqrt{2}},\cos 15°=\frac{\sqrt{3}+1}{2\sqrt{2}}]$

$\Rightarrow \cos \theta .\cos 15°+\sin \theta .\sin 15°=\frac{1}{\sqrt{2}}$

we know that,

$\Rightarrow \cos (A-B)=\cos A.\cos B+\sin A.\sin B$

and $\cos 45°=\frac{1}{\sqrt{2}}$

Using the above results we can reduce the last step as

$\Rightarrow \cos (\theta -15°)=\cos 45°$

Again, we know that the general solution of the equation $\cos \theta =\cos \alpha$ is given as, $\theta =2x\pi \pm \alpha$ $x\in Z$

$\therefore \cos (\theta -15°)=\cos 45°$

$\theta -15°=2x\pi \pm 45°$

$\theta =2x\pi +45°+15°$

$\therefore$ the general solution is

$\Rightarrow \theta =2n\pi +60°$ or $\theta =2n\pi -30°$

$\Rightarrow \theta =2n\pi +\frac{\pi }{3}$ or $\theta =2n\pi -\frac{\pi }{6}$

Hence, solved.