Let S = left[-pi, frac{pi}{2}right) - left{-f | vedclass

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(B) Given equation: $	an 	heta(1 + \sqrt{5} 	an 2	heta) = \sqrt{5} - 	an 2	heta$ $	an 	heta + \sqrt{5} 	an 	heta 	an 2	heta = \sqrt{5} - \

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$5$

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(B) Given equation: $	an 	heta(1 + \sqrt{5} 	an 2	heta) = \sqrt{5} - 	an 2	heta$ $	an 	heta + \sqrt{5} 	an 	heta 	an 2	heta = \sqrt{5} - 	an 2	heta$ $	an 	heta + 	an 2	heta = \sqrt{5}(1 - 	an 	heta 	an 2	heta)$ $\frac{	an 	heta + 	an 2	heta}{1 - 	an 	heta 	an 2	heta} = \sqrt{5}$ $	an(3	heta) = \sqrt{5}$ Let $	an \alpha = \sqrt{5}$,where $\alpha \in (0, \pi/2)$. Then $3	heta = n\pi + \alpha$,so $	heta = \frac{n\pi}{3} + \frac{\alpha}{3}$. Since $	heta \in [-\pi, \pi/2)$,we check values of $n$: For $n = -3: 	heta = -\pi + \alpha/3$ (Valid) For $n = -2: 	heta = -2\pi/3 + \alpha/3$ (Valid) For $n = -1: 	heta = -\pi/3 + \alpha/3$ (Valid) For $n = 0: 	heta = \alpha/3$ (Valid) For $n = 1: 	heta = \pi/3 + \alpha/3$ (Valid,since $\alpha/3 We must exclude values where $	an 	heta$ or $	an 2	heta$ are undefined or lead to division by zero. The set $S$ excludes $\pm \pi/2, \pm \pi/4, -3\pi/4$. All $5$ values are within the domain $S$. Thus,the number of elements is $5$.

Let $S = \left[-\pi, \frac{\pi}{2}\right) - \left\{-\frac{\pi}{2}, -\frac{\pi}{4}, -\frac{3\pi}{4}, \frac{\pi}{4}\right\}$. Then the number of elements in the set $\{\theta \in S : \tan \theta(1 + \sqrt{5} \tan(2\theta)) = \sqrt{5} - \tan(2\theta)\}$ is:

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