Question

The number of solutions of the equation $\cos \theta +\cos 3\theta +\cos 5\theta +\cos 7\theta =0,$ where $\theta \in [0,2\pi ]$ is

Answer 1

We have $\cos \theta +\cos 7\theta +\cos 3\theta +\cos 5\theta =0$
$\Rightarrow 2\cos 4\theta \cos 3\theta +2\cos 4\theta \cos \theta =0$
$\Rightarrow \cos 4\theta (\cos 3\theta +\cos \theta )=0$
$\Rightarrow \cos 4\theta \left(2\cos 2\theta \cos \theta \right)=0$
Now, either $\cos \theta =0$
$\Rightarrow \theta =(2n+1)\frac{\pi }{2},n\in I\therefore \theta =\frac{\pi }{2},\frac{3\pi }{2}$
or $\cos 2\theta =0$
$\Rightarrow \theta =(2n+1)\frac{\pi }{4},n\in I\therefore \theta =\frac{\pi }{4},\frac{3\pi }{4},\frac{5\pi }{4},\frac{7\pi }{4}$
$\text{or}\cos 4\theta =0\Rightarrow \theta =(2n+1)\frac{\pi }{8},n\in I\therefore \theta =\frac{\pi }{8},\frac{3\pi }{8},\frac{5\pi }{8},\frac{7\pi }{8},\frac{9\pi }{8},\frac{11\pi }{8},\frac{13\pi }{8},\frac{15\pi }{8}$
So, total number of solutions is $14.$