Question

Solve the following inequalities.
(i) ${\cos }^{-1}x>{\cos }^{-1}{x}^2$.
(ii) $arc{\cot }^{2}x-5arc\cot x+6>0$.

Answer 1

(i) Substitute $x=\cos y$

$\therefore {\cos }^{-1}\cos y>{\cos }^{-1}{\cos }^{2}y$

$\therefore \cos y>{\cos }^{2}y$

$\therefore y=(0,\frac{\pi }{2})\cup (\frac{3\pi }{2},2\pi )$

$\therefore x=\cos y=(\cos \frac{\pi }{2},\cos 0)\cup (\cos 2\pi ,\cos \frac{3\pi }{2})$

$\therefore x\in (0,1)$

(ii) Substitute $y={\cot }^{-1}x$

$\therefore y^2-5y+6>0$

$\therefore y^2-3y-2y+6>0$

$\therefore (y-3)(y-2)>0$

$\therefore y<2\cup y>3$

But, $y={\cot }^{-1}x$

$\therefore y\in (0,2)\cup (3,\pi )$

$\therefore x\in (-\infty ,\cot 3)\cup (\cot 2,\infty )$