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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?

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} & = & (6sqrt{2}+13)x^2 \
end{eqnarray*}
The positive solution gives:
$$boxed{x = frac{1}{sqrt{2(6sqrt{2}+13)}}}$$

There is a useful identity that we can use in this case:

$arcsin{x}+arcsin{y}=arcsin{(xsqrt{1-y^2}+ysqrt{1-x^2})}$

From here we can substitute:

$arcsin{(2xsqrt{1-9x^2}+3xsqrt{1-4x^2})}=frac{pi}{4}$

We are then left with:

$2xsqrt{1-9x^2}+3xsqrt{1-4x^2}=sin{frac{pi}{4}}$

From here, you can solve for $x$.

Let $theta_1 = sin^{-1}(2x)$,$theta_2 = sin^{-1}(3x)$. Then,
$$sin(theta_1+theta_2) = sin(theta_1)cos(theta_2)+sin(theta_2)cos(theta_1) = sin(pi/4) = 1/sqrt{2}$$
$$2xsqrt{1-9x^2}+3xsqrt{1-4x^2} = frac{1}{sqrt{2}}.$$
But I don't know how you would solve this last equation.

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
$$e^{iy}=ix+sqrt{1-x^2}$$
$$iy=lnbig[ix+sqrt{1-x^2}big]$$
$$arcsin x=-ilnbig[ix+sqrt{1-x^2}big]$$
So we look at your equation:
$$arcsin 2x+arcsin3x=fracpi4$$
$$-ilnbig[2ix+sqrt{1-4x^2}big]-ilnbig[3ix+sqrt{1-9x^2}big]=fracpi4$$
$$lnbig[2ix+sqrt{1-4x^2}big]+lnbig[3ix+sqrt{1-9x^2}big]=frac{ipi}4$$
Using the property $ln(ab)=ln a+ln b$ we see that
$$lnbigg[big(2ix+sqrt{1-4x^2}big)big(3ix+sqrt{1-9x^2}big)bigg]=frac{ipi}4$$
Taking $exp$ on both sides,
$$big(2ix+sqrt{1-4x^2}big)big(3ix+sqrt{1-9x^2}big)=e^{ipi/4}$$
Use the formula $e^{itheta}=costheta+isintheta$ 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
$$x=sqrt{frac{13}{194}-frac{3sqrt2}{97}}$$
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
$$xbigg(2sqrt{1-9x^2}+3sqrt{1-4x^2}bigg)+6ix^2-isqrt{(1-4x^2)(1-9x^2)}=frac{1-i}{sqrt{2}}$$
We set the real parts of each side equal to eachother:
$$xbigg(2sqrt{1-9x^2}+3sqrt{1-4x^2}bigg)=frac1{sqrt2}$$
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
$$y=frac{13}{194}+frac{3sqrt2}{97}$$
Taking $sqrt{cdot}$ on both sides,
$$x=sqrt{frac{13}{194}-frac{3sqrt2}{97}}$$

We need $-1le3xle1$

But if $xle0,$ the left hand side $le0$

Now $3x=sin(pi/4-arcsin(2x))$

$3sqrt2x=sqrt{1-(2x)^2}-2x$

$sqrt{1-4x^2}=x(3sqrt2+2)$

Square both sides

Start with $$2xsqrt{1-9x^2}+3xsqrt{1-4x^2} = frac{1}{sqrt{2}}$$ 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
$$72 y^2-13y+frac 12=12 ysqrt{1-9 y} sqrt{1-4 y} $$
Square again, expand and simplify to get
$$97 y^2-13 y+frac{1}{4}=0$$ which is simple.