sine/cosine=tangens, equation help.

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That's simple enough. But then the next equation pops up, and it goes like this:

sin^2 x -3 sin x * cos x + 2 cos^2 x = 0 , x between [-pi , pi]

I'm following a text book, and in the examples of how to solve this, by dividing sine with cosine, all of them had the same exponents. In this equation there are some in the first power and some in the second. I have the sneaking suspicion that this is rather simple, but I am stuck in my own thoughts, and could use some help. Smile

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@Cyracuz,

Fortunately each of your terms is one trig function multiplied by another. The first term is sin*sin; the second is sin*cos; the third is cos*cos. If you divide everything by cos^2 x and then substitute u for tan(x), you can then solve the resulting quadratic equation for u.

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@Kolyo,

That sounds like just the thing. It didn't occur to me to think of "sin^2" as one term. Thanks. If this thread dies here, take it as a good sign. Smile

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Great success. I was able to solve it with all 4 solutions to x. Thanks.

Took some time to figure it out though, when my calculator gave me the answer 0.7853.... instead of pi/4.

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@Cyracuz,

Or factor the original polynomial:

(sinx-2cosx)(sinx-cosx) = 0
sinx = 2cosx ==> tanx = 2
sinx = cosx ==> tanx = 1

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@markr,

At the moment I do not see how you did that. I see that the answers you arrived at are the right ones.

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@Cyracuz,

I would have used Mark's approach also although I like the elegance of Kolyo's method.

Quote:

sin^2 x -3 sin x * cos x + 2 cos^2 x = 0

Let A = sin X and B = cos X then you get A^2 - 3AB + 2B^2 = 0 which factors to (A-B)(A-2B) = 0.

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sine/cosine=tangens, equation help.

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