Question

Solve:

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

Answer 1

$\cos 6\theta +\cos 4\theta +\cos 2\theta +1=0$

Apply $\cos C+\cos D=2\cos \left(\frac{C+D}{2}\right)\cos \left(\frac{C-D}{2}\right)$

On $\cos 4\theta$ & $\cos 2\theta$

$\cos 6\theta +2\cos 3\theta \cos \theta +1=0$

$2{\cos }^{2}3\theta -1+2\cos 3\theta \cos \theta =-1$ $[\cos 2x=2{\cos }^{2}x-1]$

$2{\cos }^{2}3\theta +2\cos 3\theta \cos \theta =0$

$\cos 3\theta \left(\cos 3\theta \cos \theta \right)=0$

$\cos 3\theta =0$ in $0\le \theta \le \pi$

$\theta =\frac{\pi }{6},\frac{\pi }{2}[3\theta =\frac{\pi }{2},\frac{3\pi }{2}]$

$\cos \theta +\cos 3\theta =0$

$4{\cos }^3\theta -3\cos \theta +\cos \theta =0$

$4{\cos }^3\theta -2\cos \theta =0$

$\cos \theta (2{\cos }^2\theta -1)=0$

$\theta =\frac{\pi }{2},\frac{\pi }{4},\frac{3\pi }{4}$

$\theta =\frac{\pi }{6}\frac{\pi }{2},\frac{\pi }{4},\frac{3\pi }{4}$(union)