The number of solutions of sqrt{tan theta} = | vedclass

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Question

(C) Given equation: $\sqrt{	an 	heta} = 2 \sin 	heta$. For $\sqrt{	an 	heta}$ to be defined,$	an 	heta \ge 0$. Also,$2 \sin 	heta \ge 0$,so $\

Vedclass · Accepted Answer

$5$

Vedclass · Answer

(C) Given equation: $\sqrt{	an 	heta} = 2 \sin 	heta$. For $\sqrt{	an 	heta}$ to be defined,$	an 	heta \ge 0$. Also,$2 \sin 	heta \ge 0$,so $\sin 	heta \ge 0$. Thus,$	heta \in [0, \pi/2] \cup \{\pi, 2\pi\}$. Squaring both sides: $	an 	heta = 4 \sin^2 	heta$. Since $\sin^2 	heta = \frac{	an^2 	heta}{1 + 	an^2 	heta}$,we have $	an 	heta = \frac{4 	an^2 	heta}{1 + 	an^2 	heta}$. Case $1$: $	an 	heta = 0 \Rightarrow 	heta = 0, \pi, 2\pi$. Case $2$: $	an 	heta 
eq 0$,divide by $	an 	heta$: $1 = \frac{4 	an 	heta}{1 + 	an^2 	heta} \Rightarrow 1 + 	an^2 	heta = 4 	an 	heta \Rightarrow 	an^2 	heta - 4 	an 	heta + 1 = 0$. Using the quadratic formula: $	an 	heta = \frac{4 \pm \sqrt{16 - 4}}{2} = 2 \pm \sqrt{3}$. For $	an 	heta = 2 + \sqrt{3}$,$	heta = \frac{5\pi}{12}$. For $	an 	heta = 2 - \sqrt{3}$,$	heta = \frac{\pi}{12}$. Total solutions are $\{0, \pi, 2\pi, \frac{\pi}{12}, \frac{5\pi}{12}\}$,which are $5$ solutions.

The number of solutions of $\sqrt{\tan \theta} = 2 \sin \theta$ for $\theta \in [0, 2\pi]$ is equal to:

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