Chapter 5 - Logarithmic, Exponential, and Other Transcendental Functions - 5.6 Exercises - Page 372: 36

$$x = \pm 1$$

$$\eqalign{ & \arccos x = \operatorname{arcsec} x \cr & {\text{cos}}\left( {\arccos x} \right) = \cos \left( {\operatorname{arcsec} x} \right) \cr & x = \frac{1}{x} \cr & {x^2} = 1 \cr & {\text{Square root both sides}} \cr & \sqrt {{x^2}} = \sqrt 1 \cr & \left| x \right| = 1 \cr & x = \pm 1 \cr} $$