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Solve: $\arcsin{\frac{3}{5}}=\arctan{x}$, $\arcsin{\frac{3}{5}}=2\arctan{x}$

Equations involving inverse trigonometric functions are solved for the unknown. The following relation is assumed proven: $$ \theta = \arcsin{\frac{a}{c}} = \arccos{\frac{\sqrt{c^2-a^2}}{c}}, $$ which is clear from the figure below.

  1. $\arcsin{\frac{3}{5}}=\arctan{x}$ $$\begin{align} \arctan{x} &= \arcsin{\frac{3}{5}} = \arctan{\frac{3}{\sqrt{5^2-3^2}}} = \arctan{\frac{3}{4}}\\ &\Rightarrow x = \frac{3}{4}.□ \end{align}$$
  2. $\arcsin{\frac{3}{5}}=2\arctan{x}$ $$\begin{align} 2\arctan{x} &= \arcsin{\frac{3}{5}} = \arctan{\frac{3}{\sqrt{5^2-3^2}}} = \arctan{\frac{3}{4}}\\ \frac{3}{4} &= \tan{\left(2\arctan{x}\right)} = \frac{2\tan{\left(\arctan{x}\right)}}{1-\tan^2{\left(\arctan{x}\right)}} \\ &= \frac{2x}{1-x^2}\\ &\Rightarrow 0 = 3x^2 +8x -3 = (3x-1)(3x+3) \\ &\Rightarrow x = \frac{1}{3} ( x = -1 \hbox{ is extraneous}).□ \end{align}$$

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