Answer
$-1.026173162$ and $0.350035015$
Work Step by Step
a.
Step 1. Let $f(x)=x^2+sin(3x)-0.99$; we have $f'(x)=2x+3cos(3x)$.
Step 2. Graphing the function, we can identify two solutions at $x\approx-1. 0.35$ (there is a third point around $x=1.245$ where $f(x)$ is very close to zero but will never reach zero).
b.
Step 1. Use Newton’s method $x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}$ with a starting point $x_0=-1$ to estimations up to $x_3$ until $|f(x_3)|$ is very small ($\lt 10^{-8}$ in this case). We obtain one solution $x_3\approx-1.026173162$, as shown in the table.
Step 2. Repeat for the other solution with a starting point of $x_0=0.35$, calculating the estimations up to $x_2$ until $|f(x_2)|$ is very small ($\lt 10^{-9}$ in this case). We obtain another solution $x_2\approx0.350035015$, as shown in the table.
