1. Roots on an equation

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

2. Originally Posted by Punch
The equation $\displaystyle 2-cos^2\theta=Acos2\theta$ where $\displaystyle A$ is a constant, has a root $\displaystyle \theta=30^o$. Find all the other roots such that $\displaystyle \theta$ is between $\displaystyle 0^o$ and $\displaystyle 360^o$
Did you manage to find A?

3. Yes i did, A = 1.549

4. Originally Posted by Punch
Yes i did, A = 1.549
ok, then use the identity cos 2a = 2 cos^2 a -1

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

6. 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

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.

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?

8. 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,

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