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?
 A: There is a useful identity that we can use in this case:
$\arcsin{x}+\arcsin{y}=\arcsin{(x\sqrt{1-y^2}+y\sqrt{1-x^2})}$
From here we can substitute:
$\arcsin{(2x\sqrt{1-9x^2}+3x\sqrt{1-4x^2})}=\frac{\pi}{4}$
We are then left with:
$2x\sqrt{1-9x^2}+3x\sqrt{1-4x^2}=\sin{\frac{\pi}{4}}$
From here, you can solve for $x$. 
A: Let $\theta_1 = \sin^{-1}(2x)$,$\theta_2 = \sin^{-1}(3x)$.  Then,
$$\sin(\theta_1+\theta_2) = \sin(\theta_1)\cos(\theta_2)+\sin(\theta_2)\cos(\theta_1) = \sin(\pi/4) = 1/\sqrt{2}$$
$$2x\sqrt{1-9x^2}+3x\sqrt{1-4x^2} = \frac{1}{\sqrt{2}}.$$
But I don't know how you would solve this last equation.
A: I'm gonna derive the general function for $\arcsin x$ then go from there. 
Recall that 
$$\sin x=\frac{e^{ix}-e^{-ix}}{2i}$$
So if $y= \arcsin x$, then $$2ix=e^{iy}-e^{-iy}$$
Letting $u=e^{iy}$, we have
$$2ix=\frac{u^2-1}{u}$$
$$u^2-2ixu-1=0$$
Use the quadratic formula to find that 
$$u=ix+\sqrt{1-x^2}$$
Thus
$$e^{iy}=ix+\sqrt{1-x^2}$$
$$iy=\ln\big[ix+\sqrt{1-x^2}\big]$$
$$\arcsin x=-i\ln\big[ix+\sqrt{1-x^2}\big]$$
So we look at your equation:
$$\arcsin 2x+\arcsin3x=\frac\pi4$$
$$-i\ln\big[2ix+\sqrt{1-4x^2}\big]-i\ln\big[3ix+\sqrt{1-9x^2}\big]=\frac\pi4$$
$$\ln\big[2ix+\sqrt{1-4x^2}\big]+\ln\big[3ix+\sqrt{1-9x^2}\big]=\frac{i\pi}4$$
Using the property $\ln(ab)=\ln a+\ln b$ we see that 
$$\ln\bigg[\big(2ix+\sqrt{1-4x^2}\big)\big(3ix+\sqrt{1-9x^2}\big)\bigg]=\frac{i\pi}4$$
Taking $\exp$ on both sides,
$$\big(2ix+\sqrt{1-4x^2}\big)\big(3ix+\sqrt{1-9x^2}\big)=e^{i\pi/4}$$
Use the formula $e^{i\theta}=\cos\theta+i\sin\theta$ to see that 
$$\big(2ix+\sqrt{1-4x^2}\big)\big(3ix+\sqrt{1-9x^2}\big)=\frac{1+i}{\sqrt2}$$
and at this point I used Wolfram|Alpha to see that 
$$x=\sqrt{\frac{13}{194}-\frac{3\sqrt2}{97}}$$
I will update my answer once I figure out how this result is found

Edit:
Expanding out the product on the Left hand side, then multiplying both sides by $-i$, we have 
$$x\bigg(2\sqrt{1-9x^2}+3\sqrt{1-4x^2}\bigg)+6ix^2-i\sqrt{(1-4x^2)(1-9x^2)}=\frac{1-i}{\sqrt{2}}$$
We set the real parts of each side equal to eachother:
$$x\bigg(2\sqrt{1-9x^2}+3\sqrt{1-4x^2}\bigg)=\frac1{\sqrt2}$$
Which @ClaudeLeibovici showed reduced to 
$$97y^2-13y+\frac14=0$$
with $y=x^2$. Using the quadratic formula, we see that 
$$y=\frac{13+\sqrt{72}}{194}$$
which reduces to 
$$y=\frac{13}{194}+\frac{3\sqrt2}{97}$$
Taking $\sqrt{\cdot}$ on both sides, 
$$x=\sqrt{\frac{13}{194}-\frac{3\sqrt2}{97}}$$
A: 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:
$$\boxed{x = \frac{1}{\sqrt{2(6\sqrt{2}+13)}}}$$
A: We need $-1\le3x\le1$
But if $x\le0,$ the left hand side $\le0$
Now $3x=\sin(\pi/4-\arcsin(2x))$
$3\sqrt2x=\sqrt{1-(2x)^2}-2x$
$\sqrt{1-4x^2}=x(3\sqrt2+2)$
Square both sides
A: 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.
