## Precalculus (6th Edition)

Use the definitions of circular functions. A. x is a number from $(-\displaystyle \frac{\pi}{2}, \displaystyle \frac{\pi}{2})$, for which $\tan x=1.$ Since $\tan 0=0\neq 1,$ zero is not the solution ($x\neq 0)$ B. x is a number from $[0, \pi]$ for which $\cos x=0$ Since $\cos 0=1\neq$0, zero is not the solution ($x\neq 0)$ $C.$x is a number from $[-\displaystyle \frac{\pi}{2}, \displaystyle \frac{\pi}{2}]$ for which $\sin x=0$ Since $\sin 0=0$, 0 is the solution