Why are these two given expressions equivalent? $6\cos^2(\frac {x}{2})-7\cos(\frac {x}{2})+2=0$ is equivalent to the expression $(3\cos(\frac {x}{2})-2)(2\cos(\frac {x}{2})-1) = 0$
Could someone explain the steps involved in evaluating the first expression to equal the second expression?
I noticed the expression in this question here, and I was curious how this could be achieved.

The answer comes from factoring.
When you foil the two parentheticals from the second expression, their products can be seen to equal the first.
We treat $\cos(\frac {x}{2})$ much like we would the variable $x$ within any given expression.
If given the expression $2x^2-3x+1 = 0$, we are able to factor this to show that $(-2x+1)(-x+1) = 0$.
Given this particular expression: $6\cos^2(\frac {x}{2})-7\cos(\frac {x}{2})+2=0$, we are able factor this to show that $(3\cos(\frac {x}{2})-2)(2cos(\frac {x}{2})-1) = 0$.

When foiled, it is shown that:
First: $3\cos(\frac {x}{2}) \cdot (2\cos(\frac {x}{2}) = 6\cos^2(\frac {x}{2})$
Second: The product of $3\cos(\frac {x}{2}) \cdot -1 = -3\cos(\frac {x}{2})$ and the product of $2cos(\frac {x}{2}) \cdot -2 = -4cos(\frac {x}{2})$, and their sum equals $-7\cos(\frac {x}{2})$
Third: $-2 \cdot -1 = 2$
Final, we add the first, second, and third steps together, and their sums are show to be equivalent to the initial expression: $6\cos^2(\frac {x}{2})$ [first] $-7\cos(\frac {x}{2})$[second]$+2$[third]$=0$.

And the result is [Final].
