Question

$\cos 6\theta +\cos 4\theta +\cos 2\theta +1=0;0\le \theta \le \pi$.

Answer 1

$(\cos 6\theta +\cos 4\theta )+(1+\cos 2\theta )=0,0\le \theta \le \pi$
$2\cos 5\theta \cos \theta +2{\cos }^2\theta =0$
or $2\cos \theta (\cos 5\theta +\cos \theta )=0$
or $4\cos \theta \cos 2\theta \cos 3\theta =0$
$\cos \theta =0$ $\therefore \theta =(n+\frac{1}{2})\pi$
$n=0$, $\theta =\pi /2$
$\cos 2\theta =0$ $\therefore 2\theta =(n+\frac{1}{2})\pi$
$\therefore \theta =(2n+1)\pi /4$
$n=0,\theta =\pi ,n=1,\theta =3\pi /4$,
$\cos 3\theta =0$
$\therefore 3\theta =n\pi +\pi /2$
$\therefore \theta =(2n+1)\pi /6$
$n=0,\theta =\pi /6,n=1,\theta =\pi /2$,
$n=2,\theta =5\pi /6$.
$\therefore \theta ={30}^o,{45}^o,{90}^o,{135}^o,{150}^o$.