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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
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Verified Answer
The correct answer is:
3
Given, $\tan \theta=\cot 5 \theta$
$$
\begin{aligned}
& \Rightarrow \quad \tan \theta=\tan \left(\frac{\pi}{2}-5 \theta\right) \\
& \Rightarrow \quad \frac{\pi}{2}-5 \theta=n \pi+\theta \\
& \Rightarrow \quad 6 \theta=\frac{\pi}{2}-n \pi \\
& \Rightarrow \quad \theta=\frac{\pi}{12}-\frac{n \pi}{6} \\
& \text { Also, } \cos 4 \theta=\sin 2 \theta=\cos \left(\frac{\pi}{2}-2 \theta\right) \\
& \Rightarrow \quad 4 \theta=2 n \pi \pm\left(\frac{\pi}{2}-2 \theta\right)
\end{aligned}
$$
Taking positive
$$
6 \theta=2 n \pi+\frac{\pi}{2} \Rightarrow \theta=\frac{n \pi}{3}+\frac{\pi}{12}
$$
Taking negative
$$
2 \theta=2 n \pi-\frac{\pi}{2} \Rightarrow \theta=n \pi-\frac{\pi}{4}
$$
Above values of $\theta$ suggests that there are only 3 common solutions.
$$
\begin{aligned}
& \Rightarrow \quad \tan \theta=\tan \left(\frac{\pi}{2}-5 \theta\right) \\
& \Rightarrow \quad \frac{\pi}{2}-5 \theta=n \pi+\theta \\
& \Rightarrow \quad 6 \theta=\frac{\pi}{2}-n \pi \\
& \Rightarrow \quad \theta=\frac{\pi}{12}-\frac{n \pi}{6} \\
& \text { Also, } \cos 4 \theta=\sin 2 \theta=\cos \left(\frac{\pi}{2}-2 \theta\right) \\
& \Rightarrow \quad 4 \theta=2 n \pi \pm\left(\frac{\pi}{2}-2 \theta\right)
\end{aligned}
$$
Taking positive
$$
6 \theta=2 n \pi+\frac{\pi}{2} \Rightarrow \theta=\frac{n \pi}{3}+\frac{\pi}{12}
$$
Taking negative
$$
2 \theta=2 n \pi-\frac{\pi}{2} \Rightarrow \theta=n \pi-\frac{\pi}{4}
$$
Above values of $\theta$ suggests that there are only 3 common solutions.
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