Solving $\sin(x) + \ln(x) = 0$ without a calculator How would you algebraically handle
$$\sin(x) + \ln(x) = 0$$
to find the zeroes without a graphing calculator?
Here's the graph

Is approximation your only hope? Can you get an exact symbolic answer?
 A: Finding the zero algebraically without a graphing calculator is possible, but a symbolic answer is impossible (I believe), so only the algebraic part can be answered.
There are multiple ways of doing it. The most straightforward one to me is to cast it as a fixed-point iteration $x\to e^{-\sin(x)}$. The Mathematica command NestList[Exp[-Sin[#]] &, 0, 20] gives a sequence of numbers converging towards 0.57871. The first four items of the sequence are
$$
1,e^{-\sin (1)},e^{-\sin \left(e^{-\sin (1)}\right)},e^{-\sin \left(e^{-\sin \left(e^{-\sin (1)}\right)}\right)}
$$
Certainly, not all fixed-point iterations converge, but in this case, it does.
A: What you can use to approximate the solution is called the Newton Algorithm. You choose a random guess $x_0$ and then you compute
$$
x_{n+1} = x_n -\frac{f\left(x_n\right)}{f'\left(x_n\right)}
$$
As long as the expression exists (not only the case considering there's a sin, cos and log ...), it will get to a solution close to the exact solution of $f\left(x\right) = 0$. Here we have a good example as the equation has exactly one solution. Let's look at what it does. Let's take $x_0 = 1$ with
$$
x_{n+1} = x_n - \frac{\sin\left(x_n\right)+\ln\left(x_n\right)}{\cos\left(x_n\right)+1/x_n}
$$
Then
$$
x_{1} = 1-\frac{\sin\left(1\right)}{1+\cos\left(1\right)} \approx 0.454
$$
and following that
$x_{2} \approx 0.567$, $x_{3} \approx 0.579$. Only three iterations are sufficient to obtain the $0.579$ geogebra gave you. However it was because my random guess was not that far from the real solution. I chose it because $f\left(1\right)=\sin\left(1\right)>0$ and $\lim\limits_{x \rightarrow 0^{+}}f\left(x\right)=-\infty$ so guessing between $0$ and $1$ was consistent.
If you'd have taken $x_0=7.0$, it'd have taken $8$ iterations. But the method can be tricky, if you take a random guess, it can enter an infinite loop and not converge to an approximation of the exact solution
