Chapter 6 - Trigonometric Functions - 6.5 Graphs of the Tangent, Cotangent, Cosecant, and Secant Functions - 6.5 Assess Your Understanding - Page 419: 55

$\{-1,0 \}$

Step 1. Separate the original equation into two cases $e^x-1=0$ or $ln(x+2)=0$ Step 2. For $e^x-1=0$, we have $e^x=1$, thus $x=ln1=0$ Step 3. For $ln(x+2)=0$, we have $x+2=e^0=1$, thus $x=-1$ Step 4. Check, we have solutions $\{-1,0 \}$