Roots on an equation

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September 27th 2010, 03:23 AM
Punch

Roots on an equation

The equation $2-cos^2\theta=Acos2\theta$ where $A$ is a constant, has a root $\theta=30^o$ . Find all the other roots such that $\theta$ is between $0^o$ and $360^o$
September 27th 2010, 05:10 AM
mathaddict

Quote:

Originally Posted by Punch

The equation $2-cos^2\theta=Acos2\theta$ where $A$ is a constant, has a root $\theta=30^o$ . Find all the other roots such that $\theta$ is between $0^o$ and $360^o$

Did you manage to find A?
September 27th 2010, 06:26 AM
Punch

Yes i did, A = 1.549
September 28th 2010, 05:12 AM
mathaddict

Quote:

Originally Posted by Punch

Yes i did, A = 1.549

ok, then use the identity cos 2a = 2 cos^2 a -1
September 30th 2010, 04:30 AM
Punch

Quote:

Originally Posted by mathaddict

ok, then use the identity cos 2a = 2 cos^2 a -1

I guess the idea is to find the value of the constant A? How do I find the other roots after that? confused
September 30th 2010, 05:32 AM
mathaddict

Quote:

Originally Posted by Punch

I guess the idea is to find the value of the constant A? How do I find the other roots after that? confused

In your original equation,

2-cos^2 b = A cos 2b

2- cos^2 b = A (2cos ^2 b-1)

2- cos^2 b = 2A cos^2 b-A

Rearrange this: 2A cos^2 b+ cos^2 b- A- 2=0

(2A+1) cos^2 b - A -2 =0

Now you have got your A, plug it in here.
September 30th 2010, 05:38 AM
Punch

Quote:

Originally Posted by mathaddict

In your original equation,

2-cos^2 b = A cos 2b

2- cos^2 b = A (2cos ^2 b-1)

2- cos^2 b = 2A cos^2 b-A

Rearrange this: 2A cos^2 b+ cos^2 b- A- 2=0

(2A+1) cos^2 b - A -2 =0

Now you have got your A, plug it in here.

sorry, but after substituting the value of A in, what do I do?
September 30th 2010, 05:45 AM
mathaddict

Quote:

Originally Posted by Punch

sorry, but after substituting the value of A in, what do I do?

It seems like you are unsure how to solve s trigonometric equation. Review your class notes which should have covered that to get some basic understanding on this before you get back to this question.

Anyways, in case if it's the square of the cosine which is confusing you,

$\cos b =\pm \sqrt{\frac{A+2}{2A+1}}$

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