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સમીકરણ $(\sqrt 3 - 1)\sin \theta + (\sqrt 3 + 1)\cos \theta = 2$ નો વ્યાપક ઉકેલ મેળવો.
$2n\pi \pm \frac{\pi }{4} + \frac{\pi }{{12}}$
$n\pi + {( - 1)^n}\frac{\pi }{4} + \frac{\pi }{{12}}$
$2n\pi \pm \frac{\pi }{4} - \frac{\pi }{{12}}$
$n\pi + {( - 1)^n}\frac{\pi }{4} - \frac{\pi }{{12}}$
Solution
(a) Let $\sqrt 3 + 1 = r\cos \alpha $ and $\sqrt 3 – 1 = r\sin \alpha $.
Then $r = \sqrt {{{(\sqrt 3 + 1)}^2} + {{(\sqrt 3 – 1)}^2}} = 2\sqrt 2 $
$\tan \alpha = \frac{{\sqrt 3 – 1}}{{\sqrt 3 + 1}} = \frac{{1 – (1/\sqrt 3 )}}{{1 + (1/\sqrt 3 )}} = \tan \left( {\frac{\pi }{4} – \frac{\pi }{6}} \right)$
$\Rightarrow \alpha = \frac{\pi }{{12}}$
The given equation reduces to
$2\sqrt 2 \cos (\theta – \alpha ) = 2 $
$\Rightarrow \cos \left( {\theta – \frac{\pi }{{12}}} \right) = \cos \frac{\pi }{4}$
$ \Rightarrow $ $\theta – \frac{\pi }{{12}} = 2n\pi \pm \frac{\pi }{4} $
$\Rightarrow \theta = 2n\pi \pm \frac{\pi }{4} + \frac{\pi }{{12}}$.
