Question

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

Answer 1

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