The set of angles btween 0 & 2pi satisfy | vedclass

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Question

$4\, cos^2	heta - 2 \,cos	heta - 1 = 0$

$cos	heta =$$\frac{{2\sqrt 2 \, \pm \,\sqrt {8 + 16} }}{8}$  $=$$\frac{{\sqrt 2 \, \pm \,\sqrt

Vedclass · Accepted Answer

$\left\{ {\frac{\pi }{{12}}\,\,,\,\,\frac{{7\pi }}{{12}}\,\,,\,\,\frac{{17\pi }}{{12}}\,\,,\,\,\frac{{23\pi }}{{12}}} \right\}$

Vedclass · Answer

$4\, cos^2	heta - 2 \,cos	heta - 1 = 0$

$cos	heta =$$\frac{{2\sqrt 2 \, \pm \,\sqrt {8 + 16} }}{8}$  $=$$\frac{{\sqrt 2 \, \pm \,\sqrt 6 }}{4}$

$cos	heta =$ $\frac{{\sqrt 6  + \sqrt 2 }}{4}$ $ \Rightarrow \,\,	heta \,\, = \,\,\frac{\pi }{{12}}\,;\,2\pi \, - \frac{\pi }{{12}}\, = \,\frac{{23\pi }}{{12}}$

$cos	heta =$ $ - \,\frac{{\sqrt 6  - \sqrt 2 }}{4}$

$cos	heta = cos(\pi -5\pi /12) ; cos(\pi +5\pi /12)$

$	heta = 7\pi /12 ; 17\pi /12$

The set of angles btween $0$ & $2\pi $ satisfying the equation $4\, cos^2 \, \theta - 2 \sqrt 2 \, cos \,\theta - 1 = 0$ is

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