Simple trigonometric problem

[John O' Meara]

1. The problem statement, all variables and given/known data
\cos (2 \theta)=- \cos ( \theta)


2. Relevant equations



3. The attempt at a solution
In general \theta = 2n \pi +/- \alpha \mbox{ where } \cos \theta = \cos \alpha
Because \cos(2 \theta) = - \cos( \theta) \mbox{ then, } 2 \theta = 2n \pi -/+ \theta
taking 2 \theta = 2n \pi - \theta \mbox{ that implies } 3 \theta = 2n \pi \mbox{ therefore } \theta = \frac{2}{3}n \pi . The book says that theta must lie between 0 and pi/2. So I must have gone wrong. Can anyone put me right on this equation. Thanks for the help.
1. The problem statement, all variables and given/known data



2. Relevant equations



3. The attempt at a solution

[Cryxic]

This is confusing. Are you trying to show that the left-hand side is equal to the right-hand side (which I don't think is true) or are you trying to solve some other problem?

[John O' Meara]

I am trying to show that the left side equals the right, but I am not sure if I am correct. The equation itself results from solving another problem. Thanks.

[ehild]

cos(theta) = -cos(alpha) implies that either

theta = alpha +2n*pi

or

theta= (pi-alpha) +2n*pi.

ehild

[Gigasoft]

You made a mistake.

-cos\theta=cos(\pi+\theta)
2\theta_{n,+}=\left(2n+1\right)\pi+\theta_{n,+}
2\theta_{n,-}=\left(2n+1\right)\pi-\theta_{n,-}
\theta_{n,+}=\left(2n+1\right)\pi
\theta_{n,-}=\frac{\left(2n+1\right)\pi}3
There are no solutions \theta_{n,+} within the interval \left[0,\frac \pi 2\right], but the solution \theta_{0,-}=\frac \pi 3 exists.

(Edit: Yeah, what they said.)

[John O' Meara]

No, the equation I want solved is \cos (2 \theta)= - \cos( \theta). I do not know how to do it. Thanks.

[John O' Meara]

Thanks Gigasoft, and thank you all.

[HallsofIvy]

What you need is the identity cos(2\theta)= cos^2(\theta)- sin^2(\theta)= cos^2(\theta)- (1- cos^2(\theta))

cos(2\theta)= 2cos^2(\theta)- 1.

You get a quadratic equation in cos(\theta).