And so begins another day in my quest to pass Calculus I.
I have a question about factoring trigonometric expressions. I'm sorry I can't be more accurate than that, but I'm not sure of the names of the operations I need to perform.
I have a problem:
$$ -2(cos(x) – cos(2x)) = 0 $$
I divide both sides by -2:
$$cos(x) – cos(2x) = 0$$
And then transform the left hand side into a polynomial using the double angle formula:
$$1 + cos(x) – 2cos^2(x) = 0$$
However this is where I get stuck. According to Wolfram Alpha I am now supposed to factor the expression to get $-((cos(x) – 1)(1 + 2cos(x))) = 0$
Maybe I am missing something but I don't understand how to factor for that step. Could somebody explain it or link me to the names of the concepts I need to Google? Thanks.
You've gotten to $1+\cos(x)-2\cos^2(x)=0$ or, equivalently $$ 2(\cos(x))^2-\cos(x)-1=0 $$ Now, to make things easier to see, let $t=\cos(x)$. If you do, your equation now looks like $$ 2t^2-t-1=0 $$ which factors as $$ (2t+1)(t-1) = 0 $$ This can only hold if either $2t=-1$ or $t=1$. Now "unsubstitute" to get two possibilities: $$ \begin{align} 2\cos(x) &= -1&\text{or}\\ \cos(x)&=1 \end{align} $$ so, finally you need to find all $x$ for which $\cos(x)=1/2$ or $\cos(x)=1$. I'll leave that to you.