The total number of real solutions of the equ | vedclass

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(C) Let $x = 	an 	heta$. The equation becomes $	heta = 	an^{-1}(2x) - \frac{1}{2} \sin^{-1}\left(\frac{6x}{9+x^2}ight)$.Let $\alpha = \frac{1}{2

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$3$

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(C) Let $x = 	an 	heta$. The equation becomes $	heta = 	an^{-1}(2x) - \frac{1}{2} \sin^{-1}\left(\frac{6x}{9+x^2}ight)$.Let $\alpha = \frac{1}{2} \sin^{-1}\left(\frac{6x}{9+x^2}ight)$,then $\sin(2\alpha) = \frac{6x}{9+x^2}$.Using $\sin(2\alpha) = \frac{2	an \alpha}{1+	an^2 \alpha}$,we have $\frac{2	an \alpha}{1+	an^2 \alpha} = \frac{6x}{9+x^2}$.Solving for $	an \alpha$,we get $	an \alpha = \frac{x}{3}$ or $	an \alpha = \frac{3}{x}$.Case $I$: $	an \alpha = \frac{x}{3}$. Substituting into $	an(	heta + \alpha) = 2x$,we get $\frac{x + x/3}{1 - x^2/3} = 2x$.This simplifies to $\frac{4x/3}{(3-x^2)/3} = 2x \Rightarrow \frac{4x}{3-x^2} = 2x$.Either $x=0$ or $2 = 3-x^2 \Rightarrow x^2=1 \Rightarrow x = \pm 1$.For $x=0, 	heta=0$. For $x=1, 	heta=\pi/4$. For $x=-1, 	heta=-\pi/4$.Case $II$: $	an \alpha = 3/x$. Substituting into $	an(	heta + \alpha) = 2x$,we get $\frac{x + 3/x}{1 - 3} = 2x \Rightarrow \frac{x^2+3}{-2x} = 2x \Rightarrow x^2+3 = -4x^2 \Rightarrow 5x^2 = -3$,which has no real solutions.Thus,the real solutions are $	heta \in \{0, \pi/4, -\pi/4\}$,giving a total of $3$ solutions.

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