Question

The number of solution of the equation $\cos 6\theta +\cos 4\theta +\cos 2\theta +1=0$ in $[0,\pi ]$ is

Answer 1

$(\cos 6\theta +\cos 4\theta )+(\cos 2\theta +1)=0\Rightarrow 2\cos 5\theta \cos \theta +2{\cos }^2\theta =0\Rightarrow 2\cos \theta (\cos 5\theta +\cos \theta )=0\Rightarrow 4\cos \theta \cos 2\theta \cos 3\theta =0\Rightarrow \cos \theta =0\text{or}\cos 2\theta =0\text{or}\cos 3\theta =0\because \theta \in [0,\pi ]\Rightarrow \theta =\frac{\pi }{2}\text{or}\theta =\frac{\pi }{4},\frac{3\pi }{4}\text{or}\theta =\frac{\pi }{6},\frac{5\pi }{6}$
Hence number of solution$=5$