How to determine if the solutions of $$ \left(\frac{\pi}{12}+\frac{5}{3}x^2-x\sqrt{1-x}\right) = \arcsin x $$ are (compass and straightedge) constructible numbers?
I got this question from my students in my calculus class: "find the implicit derivative of " $$ \left(\frac{\pi}{12}+\frac{5}{3}x^2-x\sqrt{1-x}\right) = \arcsin x, $$ which is a simple question, but this leaves me with the curiosity of knowing if this equation really has real solutions. Then Wolfram Alpha helped me to find 2 real solutions. But now I want to know more on these numbers: are these two number (compass and straightedge) constructible? I am aware, having consulted the relevant Wikipedia entry, that
"$x$ is constructible if and only if there is a closed-form expression for $x$ using only integers and the operations for addition, subtraction, multiplication, division, and square roots."
but how can I apply this to the roots of the above equation?
The answer is no. Any solution of your equation must be transcendental, but a number constructible with a compass and ruler must be algebraic of degree that's a power of $\ 2\ .$
First note that the sum, product and ratio of any pair of algebraic numbers, and the square root of any non-negative algebraic number, must be algebraic, and \begin{align} \sin\frac{\pi}{12}&=\frac{\sqrt{3}-1}{2\sqrt{2}}\\ \cos\frac{\pi}{12}&=\frac{\sqrt{3}+1}{2\sqrt{2}}\ . \end{align} Writing your equation as $$ \frac{5x^2}{3}-x\sqrt{1-x}=\arcsin x-\frac{\pi}{12} $$ and taking the sine of both sides gives \begin{align} \sin\left(\frac{5x^2}{3}-x\sqrt{1-x}\right)&=\sin\left(\arcsin x-\frac{\pi}{12}\right)\\ &=x\cos\frac{\pi}{12}-\sin\frac{\pi}{12}\cos\arcsin x\\ &=\left(\frac{\sqrt{3}+1}{2\sqrt{2}}\right)x-\left(\frac{\sqrt{3}-1}{2\sqrt{2}}\right)\sqrt{1-x^2}\ . \end{align} Now if $\ x\ $ were algebraic, so would be $\ i\left(\frac{5x^2}{3}-x\sqrt{1-x}\right)\ ,$ $\ \left(\frac{\sqrt{3}+1}{2\sqrt{2}}\right)x-\left(\frac{\sqrt{3}-1}{2\sqrt{2}}\right)\sqrt{1-x^2}\ ,$ and hence also $\ \sin\left(\frac{5x^2}{3}-x\sqrt{1-x}\right)\ .$ Also, since either $\ \cos\left(\frac{5x^2}{3}-x\sqrt{1-x}\right)=$$\,\sqrt{1-\sin^2\left(\frac{5x^2}{3}-x\sqrt{1-x}\right)}\ $ or $\ \cos\left(\frac{5x^2}{3}-x\sqrt{1-x}\right)=$$\,{-}\sqrt{1-\sin^2\left(\frac{5x^2}{3}-x\sqrt{1-x}\right)}\ $ then $\ \cos\left(\frac{5x^2}{3}-x\sqrt{1-x}\right)\ $ would be algebraic too. Thus $$ e^{i\left(\frac{5x^2}{3}-x\sqrt{1-x}\right)}=\cos\left(\frac{5x^2}{3}-x\sqrt{1-x}\right)+i\sin\left(\frac{5x^2}{3}-x\sqrt{1-x}\right) $$ would also have to be algebraic. But the Lindemann-Weierstrass theorem tells us that if $\ i\left(\frac{5x^2}{3}-x\sqrt{1-x}\right)\ $ is algebraic and non-zero, then $\ e^{i\left(\frac{5x^2}{3}-x\sqrt{1-x}\right)}\ $ must be transcendental. It therefore follows that $\ x\ $ cannot be algebraic.