I'm studying the function
$$ f(x) = \log(x + 1) + \cos(x)/2 $$
The first derivative is:
$$ f'(x) = 1/(x + 1) − \sin(x)/2. $$
To find the first two positive critical points (without Wolfram and the like), should I be using something like the Newton–Raphson method (from a previous, similar problem: Root of Logarithmic Equation) or is there a simpler way, when dealing with trigonometric functions?
On
Newton-Raphson would be a good way to go. As an experiment, it is also possible to solve the derivative for x and use that as a formula that would (hopefully) converge on an answer. For example,
$$\frac{1}{x+1}-\frac{sin(x)}{2} = 0$$ $$\frac{1}{x+1} = \frac{sin(x)}{2}$$ $$x+1 = \frac{2}{sin(x)}$$ $$x = \frac{2}{sin(x)}-1$$
Therefore,
$$x_{n+1} = \frac{2}{sin(x_{n})}-1$$
This formula will yield approximately $1.17084752029579...$, but convergence is very slow, and it doesn't find the other roots. A slightly more sophisticated technique is preferable.
Yes, you want to use numerical methods here. There is not likely to be a "closed-form" solution to $1/(x+1) + \sin(x)/2 = 0$.