1. ## problem of equation

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

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