Inverse Trigonometric Functions: Very Difficult Problems with Solutions

Solution:
Let [tex]\alpha=\arctan 2019[/tex]. Since [tex]2019\gt 1[/tex], [tex]\alpha\in\left(\frac{\pi}{4}, \frac{\pi}{2}\right)[/tex]. For [tex]x=2019=\tan \alpha[/tex]:
[tex]\frac{2x}{1+x^2}=\frac{2\tan \alpha}{1+\tan ^2\alpha}=\sin 2\alpha[/tex]
Here [tex]2\alpha\in\left(\frac{\pi}{2}, \pi\right)[/tex], so [tex]\arcsin (\sin 2\alpha)=\pi-2\alpha[/tex], because [tex]\pi-2\alpha\in\left(0, \frac{\pi}{2}\right)[/tex] has the same sine. So [tex]f(2019)=2\alpha+\pi-2\alpha=\pi[/tex].

Solution:
The values of [tex]\arcsin [/tex] lie in [tex]\left[-\frac{\pi}{2}, \frac{\pi}{2}\right][/tex], so
[tex]-\frac{\pi}{2}\le 2\pi+\frac{\pi x}{2}\le\frac{\pi}{2}\iff -5\le x\le -3[/tex]
We also need [tex]\frac{3x+7}{x+1}\le 1[/tex]. For [tex]x\le -3[/tex] we have [tex]x+1\lt 0[/tex], so this means [tex]3x+7\ge x+1[/tex], i.e. [tex]x\ge -3[/tex]. So the only possible value is [tex]x=-3[/tex]. Check: [tex]\arcsin \frac{-2}{-2}=\arcsin 1=\frac{\pi}{2}[/tex] and [tex]2\pi-\frac{3\pi}{2}=\frac{\pi}{2}[/tex]. The root is [tex]x=-3[/tex].

Solution:
Let [tex]\alpha=\arccos x[/tex], so [tex]\cos\alpha=x[/tex]. For [tex]x\in\left[-\frac{1}{2}, \frac{1}{2}\right][/tex] we have [tex]\alpha\in\left[\frac{\pi}{3}, \frac{2\pi}{3}\right][/tex], so [tex]3\alpha-\pi\in\left[0, \pi\right][/tex]. By the triple-angle formula
[tex]\cos(3\alpha-\pi)=-\cos 3\alpha=-\left(4x^3-3x\right)=3x-4x^3[/tex]
An angle in [tex]\left[0, \pi\right][/tex] with cosine [tex]3x-4x^3[/tex] equals [tex]\arccos\left(3x-4x^3\right)[/tex]. So [tex]3\arccos x-\pi=\arccos\left(3x-4x^3\right)[/tex], which is the required equality.

Solution:
Note that
[tex]\frac{1}{2k^2}=\frac{(2k+1)-(2k-1)}{1+(2k+1)(2k-1)}[/tex]
By the formula [tex]\arctan a-\arctan b=\arctan \frac{a-b}{1+ab}[/tex] (valid for [tex]a\gt 0[/tex], [tex]b\gt 0[/tex]) we get
[tex]\arctan \frac{1}{2k^2}=\arctan (2k+1)-\arctan (2k-1)[/tex]
Adding these for [tex]k=1, 2, \dots, n[/tex], all middle terms cancel, and the sum is [tex]\arctan (2n+1)-\arctan 1[/tex]. Then
[tex]\tan \left(\arctan (2n+1)-\frac{\pi}{4}\right)=\frac{(2n+1)-1}{1+(2n+1)}=\frac{2n}{2n+2}=\frac{n}{n+1}[/tex]

Solution:
Let [tex]y[/tex] be the common value. Then [tex]y\in\left[0, \pi\right][/tex] and [tex]y\in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)[/tex], so [tex]y\in\left[0, \frac{\pi}{2}\right)[/tex] and [tex]\tan y=5x\ge 0[/tex], i.e. [tex]x\ge 0[/tex]. From [tex]\cos y=2x[/tex] and [tex]\tan y=5x[/tex] we get [tex]\sin y=\tan y\cos y=10x^2[/tex]. Then [tex]\sin ^2 y+\cos^2 y=1[/tex] gives
[tex]100x^4+4x^2-1=0\;\Rightarrow\;x^2=\frac{\sqrt{26}-1}{50}[/tex]
(the other root is negative). For this [tex]x\gt 0[/tex] the angle [tex]y=\arccos 2x\in\left(0, \frac{\pi}{2}\right)[/tex] has [tex]\tan y=5x[/tex], so [tex]x=\sqrt{\frac{\sqrt{26}-1}{50}}[/tex] is the only root.

Solution:
The domain is [tex]-\frac{1}{2}\le x\le\frac{1}{2}[/tex]. At [tex]x=\frac{1}{2}[/tex] we get [tex]\arctan \frac{5}{2}\lt\frac{\pi}{2}[/tex], which is true; at [tex]x=-\frac{1}{2}[/tex] we get [tex]\arctan \left(-\frac{5}{2}\right)\lt-\frac{\pi}{2}[/tex], which is false. For [tex]|x|\lt\frac{1}{2}[/tex] both sides are in [tex]\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)[/tex], where [tex]\tan [/tex] increases, so the inequality is equivalent to
[tex]5x\lt\tan (\arcsin 2x)=\frac{2x}{\sqrt{1-4x^2}}[/tex]
At [tex]x=0[/tex] both sides are equal. For [tex]x\gt 0[/tex]: [tex]\sqrt{1-4x^2}\lt\frac{2}{5}[/tex], so [tex]x^2\gt\frac{21}{100}[/tex] and [tex]x\gt\frac{\sqrt{21}}{10}[/tex]. For [tex]x\lt 0[/tex] the sign turns: [tex]\sqrt{1-4x^2}\gt\frac{2}{5}[/tex], so [tex]-\frac{\sqrt{21}}{10}\lt x\lt 0[/tex].