Question 3

Let $t = an heta$. The given equation is: $\theta = \tan^{-1}(2t) - \frac{1}{2} \sin^{-1} \left( \frac{6t}{9 + t^2} \right)$ We can rewrite the term inside $\sin^{-1}$ as: $\frac{6t}{9+t^2} = \frac{2(\frac{t}{3})}{1 + (\frac{t}{3})^2}$ Using the property $\sin^{-1} \frac{2x}{1+x^2} = 2 \tan^{-1} x$ for $|x| \le 1$, and since $\tan \theta$ range implies $t$ can be any real number, we first check $|\frac{t}{3}| \le 1$. For $|t| \le 3$, the equation becomes: $\theta = \tan^{-1}(2t) - \frac{1}{2} (2 \tan^{-1}(t/3)) = \tan^{-1}(2t) - \tan^{-1}(t/3)$ Taking tan on both sides: $\tan \theta = \tan(\tan^{-1}(2t) - \tan^{-1}(t/3))$ $t = \frac{2t - t/3}{1 + (2t)(t/3)} = \frac{5t/3}{1 + 2t^2/3} = \frac{5t}{3 + 2t^2}$ $t(3 + 2t^2) = 5t \Rightarrow t(3 + 2t^2 - 5) = 0 \Rightarrow t(2t^2 - 2) = 0$ This gives $t = 0, 1, -1$. All these values satisfy $|t| \le 3$.

• If $t=0, \tan \theta = 0 \Rightarrow \theta = 0$.
• If $t=1, \tan \theta = 1 \Rightarrow \theta = \pi/4$.
• If $t=-1, \tan \theta = -1 \Rightarrow \theta = -\pi/4$. If $|t| > 3$, the formula for $\sin^{-1}$ changes, leading to no real solutions (as shown by $-\cot \theta = \frac{7t}{3-2t^2}$). Thus, there are 3 real solutions.