Hello mangoqueen54,
We are given the two curves:
$$y=\sin(x)$$
$$y=1-x^2$$
And we are asked to find the points of intersection.
If we equate the two curves, we get:
$$\sin(x)=1-x^2$$
Which we can arrange as:
$$\sin(x)+x^2-1=0$$
So, if we define:
$$f(x)=\sin(x)+x^2-1$$
We may then find its roots. We will need to use a numeric root finding method since we cannot explicitly solve for $x$. So, we will use Newton's method:
$$x_{n+1}=x_{n}-\frac{f\left(x_{n}\right)}{f'\left(n_{n}\right)}$$
Using our definition of $f$, we obtain:
$$x_{n+1}=x_{n}-\frac{\sin\left(x_{n}\right)+x_{n}^2-1}{\cos\left(x_{n}\right)+2x_{n}}=\frac{x_{n}\cos\left(x_{n}\right)-\sin\left(x_{n}\right)+x_{n}^2+1}{\cos\left(x_{n}\right)+2x_{n}}$$
Next, let's look at a plot of our function $f$ and see how many roots there are and get a rough estimate of their values:
View attachment 2453
Now, ignoring for the moment the root approximations shown on the graph, let's just say we see that the smaller root is about -1.5 and then use Newton's recursive method to get an accurate approximate of this smaller root
$$x_0=-1.5$$
$$x_1\approx-1.41379912599863$$
$$x_2\approx-1.40963375165233$$
$$x_3\approx-1.40962400405597$$
$$x_4\approx-1.40962400400260$$
$$x_5\approx-1.40962400400260$$
Our last two successive approximations agree to 15 digits, so let's now find the other root, which we see is near x=0.5:
$$x_0=0.5$$
$$x_1\approx0.644107890053782$$
$$x_2\approx0.636750907010919$$
$$x_3\approx0.636732650918014$$
$$x_4\approx0.636732650805282$$
$$x_5\approx0.636732650805282$$
Our last two successive approximations agree to 15 digits, so we now have the two root approximations:
$$x\approx-1.40962400400260,\,0.636732650805282$$
Let's now verify that they are close approximations:
$$\sin(-1.40962400400260)\approx-0.987039832660$$
$$1-(-1.40962400400260)^2\approx-0.987039832660$$
$$\sin(0.636732650805282)\approx0.594571531399$$
$$1-(0.636732650805282)^2\approx0.594571531399$$
And so, using 12 decimal places of accuracy, we have found the points of intersection may be approximated by:
$$\bbox[10px,border:2px solid #207498]{(x,y)\approx(-1.409624004003,-0.987039832660),\,(0.636732650805,0.594571531399)}$$
