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$6\cos^2(\frac {x}{2})-7\cos(\frac {x}{2})+2=0$ 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.

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it's just multiplying out the terms in the second expression – mm-aops May 11 '14 at 14:46

3 Answers 3

up vote 1 down vote accepted

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].

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Key Idea: Knowing the roots of a polynomial let us factor it. So if $r_1, r_2,\ldots,r_n$ are all the roots of a given polynomial, then we can factorize it as: $a(x-r_1)(x-r_2)\cdots(x-r_n)$, where $a$ is the coefficient of the term with $x^n$.

Notice: $$6\,\color{red}{\cos\left(\tfrac x2\right)}^2+7\,\color{red}{\cos\left(\tfrac x2\right)}+2=0.$$ So we can get a simple quadratic equation by letting $t=\cos\left(\tfrac x2\right)$, then our equation becomes: $$6t^2-7t+2=0.\tag{$\star$}$$ Using the quadratic formula we get: $$6t^2-7t+2=0\iff t=\dfrac23\quad\color{grey}{\text{or}}\quad t=\dfrac12.$$ So we can write $(\star)$ as: $$6\left(t-\dfrac23\right)\left(t-\dfrac12\right)=0\iff (3t-2)(2t-1)=0.$$ Now replace $t$ with our original substitution.

Side note: This $t$-substitution isn't necessary at all, when you noticed that you had an expression of the form $a(\text{something})^2+b(\text{something})+c=0$ you can directly use the quadratic formula to solve for that $\text{something}$.

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Eloquently put. – Alex May 11 '14 at 14:53
Thanks @alx $\overset{\cdot\cdot}\smile$ – Hakim May 11 '14 at 14:54

We need find $pq,$ such that $pq=6\cdot2=12; p+q=7$


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