Question:

Let $t = \tan\theta$. The equation becomes: $$\theta = \tan^{-1}(2t) - \dfrac{1}{2}\sin^{-1}\left(\dfrac{6t}{9+t^2}\right)$$ Rewrite the argument of $\sin^{-1}$: $$\dfrac{6t}{9+t^2} = \dfrac{2(t/3)}{1+(t/3)^2}$$ Let $t/3 = \tan\phi$, where $\phi = \tan^{-1}(t/3) \in (-\pi/2, \pi/2)$. Then the argument is $\sin 2\phi$. Since $\phi \in (-\pi/2, \pi/2)$, $2\phi \in (-\pi, \pi)$. We define the expression piecewise: - For $2\phi \in [-\pi/2, \pi/2] \implies t \in [-3, 3]$: $$\dfrac{1}{2}\sin^{-1}(\sin 2\phi) = \phi = \tan^{-1}(t/3)$$ - For $t > 3$: $$\dfrac{1}{2}\sin^{-1}(\sin 2\phi) = \dfrac{\pi}{2} - \tan^{-1}(t/3)$$ - For $t < -3$: $$\dfrac{1}{2}\sin^{-1}(\sin 2\phi) = -\dfrac{\pi}{2} - \tan^{-1}(t/3)$$ Let us solve Case 1 ($t \in [-3, 3]$): $$\theta = \tan^{-1}(2t) - \tan^{-1}(t/3)$$ Taking the tangent of both sides: $$t = \tan(\tan^{-1}(2t) - \tan^{-1}(t/3)) = \dfrac{2t - t/3}{1 + 2t^2/3} = \dfrac{5t}{3+2t^2}$$ $$t\left(1 - \dfrac{5}{3+2t^2}\right) = 0 \implies t(2t^2 - 2) = 0 \implies t = 0, \pm 1$$ - $t = 0 \implies \theta = 0$ - $t = 1 \implies \theta = \pi/4$ - $t = -1 \implies \theta = -\pi/4$ All three values lie within the range $(-\pi/2, \pi/2)$ and satisfy the equation. Thus, we have 3 solutions. For Case 2 ($t > 3$) and Case 3 ($t < -3$): Taking tangent leads to $5t^2 = -3$, which has no real solutions. Therefore, the total number of real solutions is 3.