cot(x) + 3cot(2x) - 1 = 0

Hello this question looked straight forward enough, but I'm stuck and would really appreciate some help.

Solve on the interval $\displaystyle 0\leq x \leq\2\pi$

$\displaystyle \cot\theta + 3\cot2\theta-1=0$

I tried writing it in terms of $\displaystyle \cos\theta$ and $\displaystyle \sin\theta$

And got this;

$\displaystyle \dfrac{5\cos^2\theat - 3\sin^2\theta}{2\sin\theta\cos\theta} =1 $, among various similar equations.

I have worked on from here a bit, but can't seem to get it in terms of a trigonometric equation I can then solve. I'd really appreciate it if somone would tell me if I'm on the right track and should keep going or if I need to approach the question in a different way.

Thank you very much.

Re: cot(x) + 3cot(2x) - 1 = 0

Quote:

Originally Posted by

**Furyan** Hello this question looked straight forward enough, but I'm stuck and would really appreciate some help.

Solve on the interval $\displaystyle 0\leq x \leq\2\pi$

$\displaystyle \cot\theta + 3\cot2\theta-1=0$

I tried writing it in terms of $\displaystyle \cos\theta$ and $\displaystyle \sin\theta$

And got this;

$\displaystyle \dfrac{5\cos^2\theat - 3\sin^2\theta}{2\sin\theta\cos\theta} =1 $, among various similar equations.

I have worked on from here a bit, but can't seem to get it in terms of a trigonometric equation I can then solve. I'd really appreciate it if somone would tell me if I'm on the right track and should keep going or if I need to approach the question in a different way.

Thank you very much.

Hey Furyan! :)

How about writing it in terms of $\displaystyle \cos 2\theta$ and $\displaystyle \sin 2\theta$?

Re: cot(x) + 3cot(2x) - 1 = 0

Hi ILikeSerena,

Thanks, I'll try that:)

Re: cot(x) + 3cot(2x) - 1 = 0

Hi ILikeSerena:)

Thank you very much indeed, that worked! I would never have gone that way if you hadn't suggested it. Although in the end it was simple, it took me rather a long time to get there.

I got:

$\displaystyle 4\cos2\theta - \sin2\theta = -1$

Then I used:

$\displaystyle 4\cos2\theta - \sin2\theta \equiv R\cos(2\theta + \alpha)$

And ended up with:

$\displaystyle \sqrt{17}\cos(2\theta + \arctan\dfrac{1}{4}) = -1$

(Party)

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Re: cot(x) + 3cot(2x) - 1 = 0

Re: cot(x) + 3cot(2x) - 1 = 0

Thank you very much indeed, that worked! I would never have gone that way if you hadn't suggested it.

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