Question

The general solution of the equation $ (\sqrt{3}-1)\sin \theta +(\sqrt{3}+1)\cos \theta =2 $ is

• A. $ 2n\pi \pm \frac{\pi }{4}+\frac{\pi }{12} $
• B. $ n\pi +{{(-1)}^{n}}\frac{\pi }{4}+\frac{\pi }{12} $
• C. $ 2n\pi \pm \frac{\pi }{4}-\frac{\pi }{12} $
• D. $ n\pi +{{(-1)}^{n}}\frac{\pi }{4}-\frac{\pi }{12} $

Answer 1

Answer: (A)

Let $ \sqrt{3}+1=r\cos \alpha $ and $ \sqrt{3}-1=r\sin \alpha $ $ \therefore $ $ {{r}^{2}}{{\cos }^{2}}\alpha +{{r}^{2}}{{\sin }^{2}}\alpha ={{(\sqrt{3}+1)}^{2}}+{{(\sqrt{3}-1)}^{2}} $
$ \Rightarrow $ $ {{r}^{2}}=3+1+2\sqrt{3}+3+1-2\sqrt{3} $
$ \Rightarrow $ $ {{r}^{2}}=8 $
$ \Rightarrow $ $ r=2\sqrt{2} $ and $ \tan \alpha =\frac{r\sin \alpha }{r\cos \alpha }=\frac{\sqrt{3}-1}{\sqrt{3}+1} $
$ =\frac{1-1/\sqrt{3}}{1+1/\sqrt{3}} $
$ =\frac{\tan \frac{\pi }{4}-\tan \frac{\pi }{6}}{1+\tan \frac{\pi }{4}.\tan \frac{\pi }{6}} $ $ =\tan \left( \frac{\pi }{4}-\frac{\pi }{6} \right) $
$ \Rightarrow $ $ \tan \alpha =\tan (\pi /12) $ $ \Rightarrow $ $ \alpha =\frac{\pi }{12} $ Given, equation is $ (\sqrt{3}-1)\sin \theta +(\sqrt{3}+1)\cos \theta =2 $
$ \Rightarrow $ $ r\sin \alpha \sin \theta +r\cos \alpha \cos \theta =2 $
$ \Rightarrow $ $ 2\sqrt{2}\cos (\theta -\alpha )=2 $
$ \Rightarrow $ $ \cos \left( \theta -\frac{\pi }{12} \right)=\frac{2}{2\sqrt{2}}=\cos \left( \frac{\pi }{4} \right) $
$ \Rightarrow $ $ \theta -\frac{\pi }{12}=2n\pi \pm \frac{\pi }{4} $
$ \Rightarrow $ $ \theta =2n\pi \pm \frac{\pi }{4}+\frac{\pi }{12} $