Question

Find all values of $\theta$ between $0^o$ and ${180}^o$ satisfying the equation : $\cos 6\theta +\cos 4\theta +\cos 2\theta +1=0$

Answer 1

$cos6\theta +cos4\theta +cos2\theta +1=0$

$\Rightarrow 2cos4\theta cos2\theta +cos4\theta +1=0$

$\Rightarrow 2cos4\theta cos2\theta +2co{s}^{2}2\theta =0$

$\Rightarrow 2cos2\theta (cos4\theta +cos2\theta )=0$

$\Rightarrow 2cos2\theta \times 2cos3\theta \times cos\theta =0$

$\therefore cos\theta =0,cos2\theta =0,cos3\theta =0$

$\theta =(2n+1)\frac{\pi }{2}$ or $\theta =(2n+1)\frac{\pi }{4}$ or $\theta =(2n+1)$