Find the minimum positive noninteger root of $ \sin x=\sin \lfloor x \rfloor $. F. Petrov
Problem
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Tags: trigonometry, floor function, calculus, integration, algebra proposed, algebra
joybangla
17.04.2014 10:47
The solutions are given by $x=n\pi+(-1)^n\lfloor{x}\rfloor$ . Now let us consider two cases. Case-1 $n$ is even. then $x=n\pi+\lfloor{x}\rfloor\implies \{x\}=n\pi>1\;\forall n$ now this is impossible. Case-2 $n$ is odd. then $x=n\pi-\lfloor{x}\rfloor\implies 2\lfloor{x}\rfloor+\{x\}=n\pi$. Now we want $x$ to be as small as possible. also see $\lfloor{n\pi}\rfloor=2\lfloor{x}\rfloor$ so we have integral part is even. So the smallest candidate working is $n=9$. Giving $\lfloor{n\pi}\rfloor=28=2\lfloor{x}\rfloor$ so $x=14+9\pi-28=9\pi-14$. I hope the solution is free of calculation mistakes.
Zorger74
27.12.2020 21:50
If $\sin a=\sin b$, then either $a-b=2k\pi$ or $a+b=(2j+1)\pi$ where $k,j\in\mathbb{Z}$, which can be seen using the unit circle. The non-integer condition implies that $x-\lfloor x\rfloor\neq0$, and we know $x-\lfloor x\rfloor<1$, so this eliminates the first case.
For the second case, we must have $2\lfloor x\rfloor+\{x\}=(2j+1)\pi$, which implies that $(2j+1)\pi$ has an even integer part. Since an odd number times and odd number is odd, we want the fractional part times an odd number to be 1, which, because the fractional part of $\pi$ is less than $\frac{1}{7}$, this occurs for the first time when $(2j+1)=9$. Thus, we have $2\lfloor x\rfloor+\{x\}=9\pi$, which gives $\boxed{x=9\pi-28}$.