Question

The number of elements in the set $S=\{\theta \in [-4\pi ,4\pi ]:3{\cos }^{2}2\theta +6\cos 2\theta -$ $10{\cos }^2\theta +5=0\}$ is _________

Answer 1

Answer: 32

$3{\cos }^{2}2\theta +6\cos 2\theta -10{\cos }^2\theta +5=0$
$3{\cos }^{2}2\theta +6\cos 2\theta -5(1+\cos 2\theta )+5=0$
$3{\cos }^{2}2\theta +\cos 2\theta =0$
$\mathrm{Cos}2\theta =0$ OR $\cos 2\theta =-1/3$
$\theta \in [-4\pi ,4\pi ]$
$2\theta =(2n+1)\cdot \frac{\pi }{2}$
$\therefore \theta =\pm \pi /4.\pm 3\pi /4\dots \dots \dots \pm 15\pi /4$
Similarily $\cos 2\theta =-1/3$ gives $16$ solution