$\arccos (x) +\arccos (2x)=\frac{3\pi}{4}$
$\implies \arccos (x)=\frac{3\pi}{4}-\arccos (2x)$
$\implies \cos(\arccos (x))=\cos(\frac{3\pi}{4}-\arccos (2x))$
$\implies x=\cos(\frac{3\pi}{4})\cos(\arccos (2x))+\sin(\frac{3\pi}{4})\sin(\arccos(2x))$
$\implies x=\cos(\frac{3\pi}{4})\cos(\arccos (2x))+\sin(\frac{3\pi}{4})\sin(\arcsin(\sqrt{1-4x^2}))$
$\implies x=-\frac{2x}{\sqrt{2}}+\frac{\sqrt{1-4x^2}}{\sqrt{2}}$
$\implies \sqrt{2}x=-2x+\sqrt{1-4x^2}$
$\implies x(\sqrt{2}+2)=\sqrt{1-4x^2}$
$\implies x^2(2+4\sqrt{2}+4)=1-4x^2$
$\implies x^2(4\sqrt{2}+10)=1$
$\implies x^2=\frac{1}{4\sqrt{2}+10}$
$x=\pm\sqrt{\frac{1}{4\sqrt{2}+10}}$
But $x=-\sqrt{\frac{1}{4\sqrt{2}+10}}$ does not satisfy the given equation.
$\therefore x= \sqrt{\frac{1}{4\sqrt{2}+10}}$
I am looking for other methods to solve this equation. I applied the formula to combine two $\arccos(x)$ functions into one $\arccos(x)$ function and on simplifying I obtained a complicated equation.
$\endgroup$2 Answers
$\begingroup$$$\arccos(y)>\dfrac\pi2$$ for $y<0$
Here what if $x<0\iff2x<0$
$\endgroup$ $\begingroup$use that $$\arccos(x)+\arccos(y)=\arccos(xy-\sqrt{1-x^2}\sqrt{1-y^2})$$ for $$x+y\geq 0$$ and $$2\pi -\arccos(xy-\sqrt{1-x^2}\sqrt{1-y^2})$$ for $$x+y<0$$
$\endgroup$ 2
