# Solve the equation $2\arcsin x=\arcsin(\frac{3}{4}x)$

$$2\arcsin x=\arcsin(\frac{3}{4}x)$$ so $x\in[-1,1]$ so we have:

$2\arcsin x=y\Rightarrow\sin\frac{y}{2}=x$ and $\arcsin x=y \Rightarrow \sin y=\frac{3}{4}x\Rightarrow\frac{4}{3}\sin y=x$ , $y\in[-\frac{\pi}{2},\frac{\pi}{2}]$

$$\sin\frac{y}{2}-\frac{4}{3}\sin y=0$$$$\sin\frac{y}{2}-\frac{4}{3}\cdot 2\sin\frac{y}{2}\cos\frac{y}{2}=0$$ $$\sin\frac{y}{2}\cdot (1-\frac{8}{3}\cos\frac{y}{2})=0$$ Is it done properly at this point?

1.$\sin \frac{y}{2}=0\Rightarrow x=0$

And what else?

$\sin\frac{y}{2}\cdot (1-\frac{8}{3}\cos\frac{y}{2})=0$

$\sin\frac{y}{2} = 0$

So far so good

contining... suppose $\sin\frac{y}{2} \ne 0$

$(1-\frac{8}{3}\cos\frac{y}{2})=0\\ \frac{8}{3}\cos\frac{y}{2})=1\\ \cos\frac{y}{2}=\frac 38\\ \frac y2=\arccos\frac 38\\ x = \sin(\arccos\frac 38)\\ x = \sqrt {1 - \frac {9}{64}}\\ x = \frac {\sqrt{55}}{8}$

However, if this were the case.

since $\frac {\sqrt 2}{2}< \frac {\sqrt {55}}{8}< 1 \implies \frac {\pi}{4} < \arcsin \frac {\sqrt {55}}{8}< \frac {\pi}{2}\\2\arcsin \frac {\sqrt {55}}{8} > \frac {\pi}{2}$

while $\arcsin \frac 34 \frac {\sqrt {55}}{8} < \frac {\pi}{2}$

$x = 0$ is the only solution.

One more consideration. Had we found a non-zero solution, then $-x$ would also be a solution, and the technique here managed to obliterate that possibility.

Cutting out all of the algebra:

$\forall x\ne 0, |2\arcsin x| > |\arcsin(\frac 34 x)|$

Which means that $2\arcsin x \ne \arcsin(\frac 34 x)$ unless $x = 0$

If $2 \arcsin(x) = \arcsin(3x/4)$, we apply $\sin$ to both sides, and use the double-angle formula and the fact that $\cos(\arcsin(x)) = \sqrt{1-x^2}$ to get

$$2 x \sqrt{1-x^2} = 3x/4$$

thus either $x=0$ or $\sqrt{1-x^2} = 3/8$. Squaring both sides of the latter, $1 - x^2 = 9/64$, so $x = \pm \sqrt{55}/8$. Note that each of these is in the correct interval for $\arcsin(x)$ and $\arcsin(3x/4)$ to be defined. But they are not solutions, because $2 \arcsin(\pm \sqrt{55}/8) \approx 2.372799104$ is outside of the interval $[-\pi/2, \pi/2]$ of allowed values for arcsin. So the only correct solution is $x=0$.

HINT:

Take the sine of both sides, apply the double angle formula for the sine function, and use $\cos(\arcsin(x))=\sqrt{1-x^2}$.

• The OP has clearly taken the sine of both sides and used the double angle formula. They asked "Is it done properly at this point?" and you haven't address their actual question. – Ian Miller Jan 25 '17 at 22:58
• This also introduces two extraneous solutions $\left(x=\pm\frac{\sqrt{55}}{8}\right)$ that then need to be then be eliminated.. – Ian Miller Jan 25 '17 at 23:16
• @IanMiller No, that's not accurate at all. And why would you down vote this? – Mark Viola Jan 25 '17 at 23:22