How to sketch $\ln|\sin x|$

Question

The question is to find the number of solutions of $$\ln|\sin x|=-x^2+2x$$ but I just don't know how to sketch a logarithmic function like $\ln|\sin x|$.

A step by step method would be highly appreciated. Thanks in advance.

Answer 1

We are looking for solutions to $\sin{x} = \pm e^{-x^{2}+2x}$.

Now, (just taking one part of the right hand side), the graph of $e^{-x^{2}+2x}$ approaches $0$ as $x \to \infty$, so the graph of $y=\sin{x}$ will intersect the graph of $y=e^{-x^{2}+2x}$ infinitely many times. This will give infinitely many solutions to $\ln{|\sin{x}|} = -x^{2}+2x$.

Answer 2

First you draw $y=|\sin x|$ which is nothing else that the graph of the sine with negative part mirrored wrt $x$-axis.

Then remember that sine function is less than $1$. Logarithm of numbers in $(0,1)$ are negative ($\ln 1=0$ so at $x=\pi/2$ logarithm is zero), and that $\ln x\to -\infty$ as $x\to 0^+$ and get the graph of $y=\ln|\sin x|$.

Finally draw the parabola $y=-x^2+2x$ and see where the two curves intersect in the interval $[0,\pi]$:$x_1\approx 2$ and $x_2\approx 3$.

There are infinitely many solutions $$2 x-x^2=\log (\left| \sin x\right| )$$ $$|\sin x |=e^{2x-x^2}$$ $$\sin x=\pm e^{2x-x^2}$$

For the limitation on $\sin$ we must have $e^{2 x-x^2}\leq 1\to 2x-x^2\leq 0\to x\le 0\lor x\geq 2$

---

$$...$$

Answer 3

If you use inspection, you would notice that, if $$f(x)=\log (|\sin (x)|)+x^2-2 x$$ $$f\left(\frac{2 \pi }{3}\right)=\log \left(\frac{\sqrt{3}}{2}\right)+\frac{4 \pi ^2}{9}-\frac{4 \pi }{3}$$ is very small $(\approx 0.054)$ while $$f\left(\frac{ \pi }{2}\right)=\frac{1}{4} (\pi -4) \pi \approx -0.674$$ $$f\left(\frac{3 \pi }{4}\right)=\frac{1}{16} \left(-24 \pi +9 \pi ^2-8 \log (2)\right) \approx 0.493$$

On the other hand, there is a problem on each side of $x=n\pi$ because of the logarithm.

So, an infinite number of roots $(\sim 2, n\pi-\epsilon,n\pi+\epsilon)$