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સમીકરણ $(\sqrt 3 - 1)\,\sin \,\theta \, + \,(\sqrt 3 + 1)\,\cos \theta \, = \,2$ ના બધા $n \in Z$ ના વ્યાપક ઉકેલ મેળવો.
$\theta \, = \,2n\pi \, \pm \,\frac{\pi }{4}\, + \,\frac{\pi }{{12}}$
$\theta \, = \,n\pi \, + {( - 1)^\pi }\,\frac{\pi }{4}\, + \,\frac{\pi }{{12}}$
$\theta \, = \,2n\pi \, \pm \,\frac{\pi }{4}\, - \,\frac{\pi }{{12}}$
$\theta \, = \,n\pi \, + {( - 1)^\pi }\,\frac{\pi }{4}\, - \,\frac{\pi }{{12}}$
Solution
$(\sqrt{3}-1) \sin \theta+(\sqrt{3}+1) \cos \theta=2$
$\Rightarrow \frac{(\sqrt{3}-1)}{2 \sqrt{2}} \sin \theta+\left(\frac{\sqrt{3}+1}{2 \sqrt{2}}\right) \cos \theta=\frac{1}{\sqrt{2}}$
$\Rightarrow \sin \frac{\pi}{12} \sin \theta+\cos \frac{\pi}{12} \cos \theta=\cos \frac{\pi}{4}$
${\Rightarrow \cos \left(\theta-\frac{\pi}{12}\right)=\cos \frac{\pi}{4}} $
${\Rightarrow \theta-\frac{\pi}{12}=2 \mathrm{n} \pi \pm \frac{\pi}{4}} $
$\quad \left[ {\therefore \cos \theta = \cos \alpha \Rightarrow \theta = 2{\rm{n}}\pi \pm \alpha } \right]$
$\Rightarrow \theta=2 n \pi \pm \frac{\pi}{4}+\frac{\pi}{12},$ $n$ is any integer.
