Question

$\cos 6\theta +\cos 4\theta +\cos 2\theta +1=0.0^o<\theta <{180}^o$.

Answer 1

$\begin{array}{c}\cos 6\theta +\cos 4\theta +\cos 2\theta +1=0\\ \Rightarrow \left(\cos 6\theta +\cos 2\theta \right)+\cos 4\theta +1=0\\ \Rightarrow \left(2\cos 4\theta .\cos 2\theta \right)+\left(\cos 4\theta +1\right)=0\\ \Rightarrow 2\cos 4\theta .\cos 2\theta +2{\cos }^{2}2\theta =0\\ \Rightarrow 2\cos 2\theta \left[\cos 4\theta +\cos 2\theta \right]=0\\ \Rightarrow 2\cos 2\theta \left[2\cos 3\theta .\cos \theta \right]=0\\ \Rightarrow \therefore 2{\cos }^2\theta =0,2\theta =\left(2n+1\right)\frac{\pi }{2},\theta =\left(2n+1\right)\frac{\pi }{4}\\ \Rightarrow or2\cos 3\theta =0,3\theta =\left(2k+1\right)\frac{\pi }{2},\theta =\left(2k+1\right)\frac{\pi }{6}\\ \Rightarrow or\cos \theta =0,\theta =\left(2P+1\right)\frac{\pi }{2}\\ Where,n,k,P\in I\\ Hence\theta =\left(2P+1\right)\frac{\pi }{2}.\end{array}$