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The number of values of $\theta$ in the interval $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ such that $\theta \neq \frac{n \pi}{5}$ for $n=0, \pm 1, \pm 2$ and $\tan \theta=\cot 5 \theta$ as well as $\sin 2 \theta=\cos 4 \theta$ is
$3$
$2$
$5$
$8$
Solution
$ \tan \theta=\cot 5 \theta $
$ \Rightarrow \cos 6 \theta=0 $
$ 4 \cos ^3 2 \theta-3 \cos 2 \theta=0 $
$ \Rightarrow \cos 2 \theta=0 \text { or } \pm \frac{\sqrt{3}}{2} $
$ \sin 2 \theta=\cos 4 \theta $
$ \Rightarrow 2 \sin ^2 2 \theta+\sin 2 \theta-1=0 $
$ 2 \sin ^2 2 \theta+2 \sin 2 \theta-\sin 2 \theta-1=0 $
$ \sin 2 \theta=-1 \text { or } \sin 2 \theta=\frac{1}{2} $
$ \cos 2 \theta=0 \text { and } \sin 2 \theta=-1 $
$ \Rightarrow 2 \theta=-\frac{\pi}{2} \Rightarrow \theta=-\frac{\pi}{4} $
$ \cos 2 \theta= \pm \frac{\sqrt{3}}{2}, \sin 2 \theta=\frac{1}{2} $
$ \Rightarrow 2 \theta=\frac{\pi}{6}, \frac{5 \pi}{6} \Rightarrow \theta=\frac{\pi}{12}, \frac{5 \pi}{12} $
$ \therefore \theta=-\frac{\pi}{4}, \frac{\pi}{12}, \frac{5 \pi}{12}$