1. ## problem of equation

let $\alpha,\beta,\gamma$ be 3 positive roots of the equation $e^x\sin x -\frac{1}{2}=0$ where $\alpha<\beta<\gamma$.
show that there is a root of the equation $f(x)=e^x\sin x +\frac{1}{2}$ in $(\alpha,\gamma)$

2. It depends on what you can use (any methods of calculus?) and the level of strictness. Basically, if you look at the graph of $e^x\sin x$, this is obvious, but the question is, how much handwaving you are allowed to use in referring to the graph.
The function $e^x\sin x$ behaves like $\sin x$ in that it is positive on $(2k\pi,(2k+1)\pi)$ and negative on $((2k+1)\pi,(2k+2)\pi)$ for $k\in\mathbb{Z}$ and it is zero in $k\pi$, $k\in\mathbb{Z}$. However, the amplitude increases exponentially. You can see the graph in WolframAlpha. So if $\alpha,\beta,\gamma$ are three positive roots of $e^x\sin x -\frac{1}{2}=0$, then there is an interval $((2k+1)\pi,(2k+2)\pi)$ either inside $(\alpha, \beta)$ or inside $(\beta,\gamma)$ (or both, if these are not consecutive roots), where $e^x\sin x$ is negative. Also, one can show that even in the first negative dip between $\pi$ and $2\pi$, the function's minimum is far less than -1/2.
It depends on what you can use (any methods of calculus?) and the level of strictness. Basically, if you look at the graph of $e^x\sin x$, this is obvious, but the question is, how much handwaving you are allowed to use in referring to the graph.
The function $e^x\sin x$ behaves like $\sin x$ in that it is positive on $(2k\pi,(2k+1)\pi)$ and negative on $((2k+1)\pi,(2k+2)\pi)$ for $k\in\mathbb{Z}$ and it is zero in $k\pi$, $k\in\mathbb{Z}$. However, the amplitude increases exponentially. You can see the graph in WolframAlpha. So if $\alpha,\beta,\gamma$ are three positive roots of $e^x\sin x -\frac{1}{2}=0$, then there is an interval $((2k+1)\pi,(2k+2)\pi)$ either inside $(\alpha, \beta)$ or inside $(\beta,\gamma)$ (or both, if these are not consecutive roots), where $e^x\sin x$ is negative. Also, one can show that even in the first negative dip between $\pi$ and $2\pi$, the function's minimum is far less than -1/2.
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