The number of values of theta in the interval | vedclass

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(A) Given $	an 	heta = \cot 5 	heta = 	an\left(\frac{\pi}{2} - 5	hetaight)$.This implies $	heta = n\pi + \frac{\pi}{2} - 5	heta$,so $6	heta

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$3$

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(A) Given $	an 	heta = \cot 5 	heta = 	an\left(\frac{\pi}{2} - 5	hetaight)$.This implies $	heta = n\pi + \frac{\pi}{2} - 5	heta$,so $6	heta = n\pi + \frac{\pi}{2}$,or $	heta = \frac{(2n+1)\pi}{12}$.Also,$\sin 2	heta = \cos 4	heta = 1 - 2\sin^2 2	heta$.Let $x = \sin 2	heta$,then $2x^2 + x - 1 = 0$,which gives $(2x-1)(x+1) = 0$.So $\sin 2	heta = \frac{1}{2}$ or $\sin 2	heta = -1$.If $\sin 2	heta = -1$,then $2	heta = -\frac{\pi}{2} \Rightarrow 	heta = -\frac{\pi}{4}$.If $\sin 2	heta = \frac{1}{2}$,then $2	heta = \frac{\pi}{6}$ or $\frac{5\pi}{6} \Rightarrow 	heta = \frac{\pi}{12}$ or $\frac{5\pi}{12}$.Checking the condition $	heta 
eq \frac{n\pi}{5}$ for $n=0, \pm 1, \pm 2$:For $	heta = -\frac{\pi}{4}$,$	heta = \frac{\pi}{12}$,and $	heta = \frac{5\pi}{12}$,none of these are equal to $0, \pm \frac{\pi}{5}, \pm \frac{2\pi}{5}$.Thus,there are $3$ such values.

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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