# Determine the numbers of solutions of equations $\sin v= \frac{v}{1964}$ and $\sin v= \log_{100} v$ .

Problem. Determine the number of solutions of each following equation

• $$\sin v= \dfrac{v}{1964}$$
• $$\sin v= \log_{100} v$$

Remark. This problem is similar to another one in Moscow Olympiad 1941.

• Remark: $$v= 0$$ is a solution of $$\sin v= \dfrac{v}{1964}$$. If $$v_{0}$$ is a solution of $$\sin v= \dfrac{v}{1964}$$ then $$-v_{0}$$ is. So we just need to determine the number of positive solutions of $$\sin v= \dfrac{v}{1964}$$. We have $$v= 1964. \sin v\leqq 1964$$ Therefore, the maximium solution of $$\sin v= \dfrac{v}{1964}$$ is not over $$1964$$. We devide $${\rm Ox}$$ (also is $${\rm Ov}$$) into $$312$$ ranges of $$2\,\pi$$ length and a range of $$1964- 312.\,2\,\pi> \pi$$ length. Based on the nature of 2 function graphs $$y= \sin x= \sin v$$ and $$y= \dfrac{x}{1964}= \dfrac{v}{1964}$$ in $$(0, 1964)$$, we can determine the number of solutions of $$\sin v= \dfrac{v}{1964}$$ as follows

• The first range $$(0, 2\,\pi)$$, then $$\sin v= \dfrac{v}{1964}$$ has only a positive solution.

• All the next ranges (except the final range), $$\sin v= \dfrac{v}{1964}$$ has two positive solutions.

• The final range $$(312.\,2\,\pi, 1964)$$, because this range has a larger length than $$\pi$$, so it 'contains' enough the part of the graph $$y= \sin v$$ on $${\rm OX}$$, therefore, $$\sin v= \dfrac{v}{1964}$$ also has two positive solutions. In conclusion, the number of solutions of $$\sin v= \dfrac{v}{1964}$$ is $$1+ 311.\,2+ 2= 625$$ .

• The equation $$\sin v= \log_{100} v$$ is harder to me because I have not studied $$\log$$ yet. The condition is $$x> 0$$ and we have $$\log_{100} v\leqq 1$$. I think we should devide $${\rm Ox}$$ (also is $${\rm Ov}$$) into many ranges as a.

• The no. of solutions to the first problem is $626$ not $625$ in the first quadrant. You will subtract one occurence of $v = 0$ only when you double the number of the solutions to include the negative ones also. So, total number of solutions are $626 \mul 2 - 1 = 1521$. You can even check this on desmos. Jun 23, 2019 at 13:22
• I'll notice that, thank you a real lot !
– user680032
Jun 23, 2019 at 13:41

$$\log_{100} v =\dfrac{\log(v)}{\log(100)}$$ so $$\log(v) \le \log(100)$$ so $$v \le 100$$.