Question

The number of solutions of the equation $\cos \theta +\cos 4\theta +\cos 7\theta =0$ in $[0,2\pi ]$ is

Answer 1

$\cos \theta +\cos 4\theta +\cos 7\theta =0\Rightarrow 2\cos 4\theta \cdot \cos 3\theta +\cos 4\theta =0\Rightarrow \cos 4\theta =0\text{or}2\cos 3\theta +1=0$

For $\cos 4\theta =0$
$4\theta =(2n+1)\frac{\pi }{2},n\in Z\Rightarrow \theta =(2n+1)\frac{\pi }{8}\Rightarrow \theta =\frac{\pi }{8},\frac{3\pi }{8},\frac{5\pi }{8},\frac{7\pi }{8},\frac{9\pi }{8},\frac{11\pi }{8},\frac{13\pi }{8},\frac{15\pi }{8}$
$8$ solutions

For $2\cos 3\theta +1=0$
$\cos 3\theta =-\frac{1}{2}\Rightarrow 3\theta =2n\pi +\frac{2\pi }{3}\text{or}2n\pi +\frac{4\pi }{3},n\in Z\Rightarrow \theta =\frac{2\pi }{9},\frac{4\pi }{9},\frac{8\pi }{9},\frac{10\pi }{9},\frac{14\pi }{9},\frac{16\pi }{9}$
6 solutions

So, there are total $(8+6)=14$ solutions.