The number of solutions of the trigonometric | vedclass

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(C) Given equation: $2 	an 2 	heta - \cot 2 	heta + 1 = 0$. Let $x = 	an 2 	heta$. Then $\cot 2 	heta = \frac{1}{x}$. The equation becomes $2x -

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$4$

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(C) Given equation: $2 	an 2 	heta - \cot 2 	heta + 1 = 0$. Let $x = 	an 2 	heta$. Then $\cot 2 	heta = \frac{1}{x}$. The equation becomes $2x - \frac{1}{x} + 1 = 0$,which simplifies to $2x^2 + x - 1 = 0$. Factoring the quadratic: $(2x - 1)(x + 1) = 0$,so $x = \frac{1}{2}$ or $x = -1$. Case $1$: $	an 2 	heta = -1$. $2 	heta = n \pi - \frac{\pi}{4} \Rightarrow 	heta = \frac{n \pi}{2} - \frac{\pi}{8}$. For $	heta \in [0, \pi]$,possible values are $	heta = \frac{3 \pi}{8}$ $(n=1)$ and $	heta = \frac{7 \pi}{8}$ $(n=2)$. Case $2$: $	an 2 	heta = \frac{1}{2}$. $2 	heta = n \pi + 	an^{-1}(\frac{1}{2}) \Rightarrow 	heta = \frac{n \pi}{2} + \frac{1}{2} 	an^{-1}(\frac{1}{2})$. For $	heta \in [0, \pi]$,possible values are $	heta = \frac{1}{2} 	an^{-1}(\frac{1}{2})$ $(n=0)$ and $	heta = \frac{\pi}{2} + \frac{1}{2} 	an^{-1}(\frac{1}{2})$ $(n=1)$. Thus,there are $2 + 2 = 4$ solutions in the interval $[0, \pi]$.

The number of solutions of the trigonometric equation $2 \tan 2 \theta - \cot 2 \theta + 1 = 0$ lying in the interval $[0, \pi]$ is

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