Chapter 6 - Inverse Circular Functions and Trigonometric Equations - Section 6.4 Equations Involving Inverse Trigonometric Functions - 6.4 Exercises - Page 280: 33

$\frac{4}{5}$

Given: $cos^{-1}x=sin^{-1}\frac{3}{5}$ Consider $sin^{-1}\frac{3}{5}=P$ $sin P=\frac{3}{5}$ Apply trigonometric identity for sine. $sinx=\frac{opp}{hyp}=\frac{3}{5}$ Thus, $adj=\sqrt {5^{2}-3^{2}}=\sqrt {25-9}=\sqrt {16}=4$ Likewise, $cosP=\frac{adj}{hyp}=\frac{4}{5}$ $cos^{-1}x=P$ gives $x=cosP$ Hence, $x=\frac{4}{5}$