<s>[INST] Solving for $x$ in $\sin^{-1}(2x) + \sin^{-1}(3x) = \frac \pi 4$ 
Given an equation: $$\sin^{-1}(2x) + \sin^{-1}(3x) = \frac \pi 4$$
How do I find $x$?

I tried solving by differentiating both sides, but I get $x=0$.
How do you solve it, purely using trigonometric techniques?
 [/INST] Or this way using


*

*$\cos(a+b) = \cos a \cos b - \sin a \sin b$

*$\cos a = \sqrt{1-\sin^2 a}$
\begin{eqnarray*}
\sin^{-1}(2x) + \sin^{-1}(3x) & = & \frac \pi 4 \\
\sqrt{1-4x^2}\sqrt{1-9x^2} - 6x^2 & = & \frac{\sqrt{2}}{2} \\
(1-4x^2)(1-9x^2) &=& \left( \frac{\sqrt{2}}{2} +6x^2 \right)^2 \\
\frac{1}{2} & = & (6\sqrt{2}+13)x^2 \\
\end{eqnarray*}
The positive solution gives:
$$\boxed{x = \frac{1}{\sqrt{2(6\sqrt{2}+13)}}}$$
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