The general solution of the equation (sqrt 3 | vedclass

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Question

(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 $

Vedclass · Accepted Answer

$2n\pi \pm \frac{\pi }{4} + \frac{\pi }{{12}}$

Vedclass · Answer

(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 $

$	an \alpha = \frac{{\sqrt 3 - 1}}{{\sqrt 3 + 1}} = \frac{{1 - (1/\sqrt 3 )}}{{1 + (1/\sqrt 3 )}} = 	an \left( {\frac{\pi }{4} - \frac{\pi }{6}} ight)$

$\Rightarrow \alpha = \frac{\pi }{{12}}$

The given equation reduces to

$2\sqrt 2 \cos (	heta - \alpha ) = 2 $

$\Rightarrow \cos \left( {	heta - \frac{\pi }{{12}}} ight) = \cos \frac{\pi }{4}$

$ \Rightarrow $ $	heta - \frac{\pi }{{12}} = 2n\pi \pm \frac{\pi }{4} $

$\Rightarrow 	heta = 2n\pi \pm \frac{\pi }{4} + \frac{\pi }{{12}}$.

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

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