<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] Start with $$2x\sqrt{1-9x^2}+3x\sqrt{1-4x^2} = \frac{1}{\sqrt{2}}$$ and square both sides to get
$$-72 x^4+13x^2+12 \sqrt{1-9 x^2} \sqrt{1-4 x^2} x^2=\frac 12$$ that is to say
$$72 x^4-13x^2+\frac 12=12 \sqrt{1-9 x^2} \sqrt{1-4 x^2} x^2$$ Let $y=x^2$ to make
$$72 y^2-13y+\frac 12=12 y\sqrt{1-9 y} \sqrt{1-4 y} $$
 Square again, expand and simplify to get
$$97 y^2-13 y+\frac{1}{4}=0$$ which is simple.
</s>