Question 3

**Solution sketch** Let $t=\tan\theta$. Note $\dfrac{6t}{9+t^2}=\sin\!\big(2\tan^{-1}\tfrac t3\big)$. **Range of RHS:** $\tan^{-1}(2t)\in(-\tfrac\pi2,\tfrac\pi2)$ and $\tfrac12\sin^{-1}(\cdot)\in[-\tfrac\pi4,\tfrac\pi4]$, so $\theta\in(-\tfrac{3\pi}{4},\tfrac{3\pi}{4})$. For $\theta\in(\tfrac\pi2,\tfrac{3\pi}4)$ we have $t<0$, making RHS $<\tfrac\pi4$; impossible (similarly on the negative side). So $\theta\in(-\tfrac\pi2,\tfrac\pi2)$, i.e. $\theta=\tan^{-1}t$. **Case $|t|\le 3$:** $\tfrac12\sin^{-1}\!\left(\dfrac{6t}{9+t^2}\right)=\tan^{-1}\tfrac t3$, so $$\theta=\tan^{-1}(2t)-\tan^{-1}\tfrac t3=\tan^{-1}\frac{5t}{3+2t^2}.$$ Hence $t=\dfrac{5t}{3+2t^2}\Rightarrow t=0,\ \pm1$, giving $\theta=0,\pm\tfrac\pi4$ (all valid). **Case $t>3$:** $\tfrac12\sin^{-1}(\cdot)=\tfrac\pi2-\tan^{-1}\tfrac t3$, so $$\tan^{-1}t=\tan^{-1}(2t)+\tan^{-1}\tfrac t3-\tfrac\pi2=\tfrac\pi2+\tan^{-1}\frac{7t}{3-2t^2}.$$ Taking tangents: $t=\dfrac{2t^2-3}{7t}\Rightarrow 5t^2=-3$, no solution. By oddness, none for $t<-3$ either. Total real solutions: $3$.

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