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I have a problem with the following trigonometric equation: $$ 3\sin(x)^2 - 2\sin(x)\cos(x) - \cos(x)^2 = 0 $$

It's from the book Engineering Mathematics 7th edition by Stroud. The book is giving the answer, but I can't seem to be able to find out how to factorize it. I can't figure it out.

Consider the following solution:

This equation can be written as $3\sin ^2x-\cos ^2x = 2\sin x \cos x.$

That is: $3\sin ^2x-2\sin x \cos x-\cos ^2 x = 0.$

That is: $(3\sin x + \cos x)(\sin x - \cos x) = 0.$

So that $3 \sin x \cos x = 0$ or $\sin x - \cos x = 0.$

If $3 \sin x + \cos x = 0,$ then $\tan x = \frac{-1}{3},$ and so $x = -0.3218 ± n \pi,$ and if $\sin x - \cos x = 0,$ then $\tan x = 1,$ and so $x = \frac{\pi}{4}.$

If anyone could help me understand how to factorize this equation to get the one shown in the image it would help me very much.

Thank you in advance.

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    $\begingroup$ Can you verify that $$(3\sin x + \cos x)(\sin x - \cos x) = 3\sin^2 x - 2\sin x\cos x -\cos^2 x$$ by expanding the product? $\endgroup$
    – AlexR
    Apr 27, 2015 at 14:09
  • $\begingroup$ Yes of course, i can do that. But how can you go from the 2nd to the 1st one, without knowing the answer. I'm trying to use the standard formulas etc but i can't find it. $\endgroup$
    – KeyC0de
    Apr 27, 2015 at 14:12
  • $\begingroup$ If you put $\sin x=\cos x$, you will find that the expression reduces to $0$, which means that $\sin x-\cos x$ is a factor. $\endgroup$
    – Apurv
    Apr 27, 2015 at 14:15
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    $\begingroup$ @RestlessC0bra: The standard trig formulas are a distraction here. Temporarily forget that it's a trig equation & convert it to an equation in $a$ and $b$ (or $u$ and $v$), as in the answers below. $\endgroup$
    – PM 2Ring
    Apr 27, 2015 at 14:31

3 Answers 3

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If you put $\sin x = a$, and $\cos x = b$, then you might be able to see the "structure" of the equation:

$$\begin{align} 3\sin^2x - 2\sin x \cos x - \cos^2x & = 3a^2 - 2ab - b^2 \\ &= 3a^2-3ab+ab-b^2\\ & =3a(a-b)+b(a-b)\\ & =(a-b)(3a+b)\\ & = (3a+ b)(a-b) \\ & = (3\sin x + \cos x)(\sin x - \cos x) = 0\end{align}$$

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  • $\begingroup$ I understand. I got confused with these trigonometric equations and then got stuck. Thanks everyone very much! $\endgroup$
    – KeyC0de
    Apr 27, 2015 at 14:42
  • $\begingroup$ You're welcome! $\endgroup$ Apr 27, 2015 at 14:45
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Think of it like this:

Take $a=\sin x$ and $b=\cos x$. Then you have,

$$3a^2-2ab-b^2=3a^2-3ab+ab-b^2=3a(a-b)+b(a-b)=(a-b)(3a+b)$$

Substitute everything back and you have,

$$3\sin^2 x-2\sin x\cos x-\cos^2 x=(\sin x-\cos x)(3\sin x+\cos x)$$

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Let $u = \sin x$; let $v = \cos x$. Then the equation $$3\sin^2x - 2\sin x\cos x - \cos^2x = 0$$ becomes $$3u^2 - 2uv - v^2 = 0$$ To split the linear term, we must find two numbers with product $3 \cdot -1 = -3$ and sum $-2$. They are $-3$ and $1$. Hence, \begin{align*} 3u^2 - 2uv - v^2 & = 0\\ 3u^2 - 3uv + uv - v^2 & = 0 && \text{split the linear term}\\ 3u(u - v) + v(u - v) & = 0 && \text{factor by grouping}\\ (3u + v)(u - v) & = 0 && \text{extract the common factor} \end{align*} Now, substitute $\sin x$ for $u$ and $\cos x$ for $v$.

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