A way for cosine of a decimal to equal to a cosine of a whole number?

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I wasn't really sure where to put this since this relates to programming as well, and I'm sorry if this question is off-topic or plain ignorant. I was hoping to find a method of having a cosine of a whole number equal to a cosine of a decimal. Here's an example

$\cos(0.8) = \cos(x), x \in \mathbb{Z}$

So that an cosine of an integer is equal to cosine of 0.8. The reason being is that I'm scripting in Batch, and it does not handle any floating point arithmetic, but I'm trying to calculate the cosine of an angle. If it helps at all, I have a cosine function that calculates the value of cosine *10000.

SET "_SIN=a-a*a/1920*a/312500+a*a/1920*a/15625*a/15625*a/2560000-a*a/1875*a/15360*a/15625*a/15625*a/16000*a/44800000"
SET "cos=(a=(15708 - (x) * 31416 / 180)%%62832, c=(a>>31|1)*a, a-=(((c-47125)>>31)+1)*((a>>31|1)*62832)  +  (-((c-47125)>>31))*( (((c-15709)>>31)+1)*(-(a>>31|1)*31416+2*a)  ), %_SIN%)"

Many thanks.

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$\cos(\alpha)=\cos(\beta)$ if and only if there is $n\in\mathbb{Z}$ such that $\alpha + \beta = 2\pi n$ or $\alpha - \beta = 2\pi n.$ Now assume $\alpha = 0.8$ and $\beta\in\mathbb{Z}.$ Then we would have $\pi = \frac{\alpha \pm\beta}{2n}\in\mathbb{Q}.$ This is a contradiction, because $\pi$ is irrational. Therefore, there is no $x\in\mathbb{Z}$ such that $\cos(0.8)=\cos(x).$