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trig help

1. The problem statement, all variables and given/known data
Solve the equation 2. Relevant equations

quadratic equation,
power reduction formula:  
3. The attempt at a solution

First I tried using the quadratic equation to find the roots of cos theta and got:and an inadmissible root.
Theta has to be between 3pi/2 and 5pi, and since I don't know the period, I wasn't sure how to express the roots generally in terms of n.
So then I tried the power reduction formula for cos^2 theta and got
but I don't know where to go from there.  Can anyone help me please?  

    You did it the right way the first time, the period of cos is always 2pi so I don't understand what's troubling you.  

    I guess what's troubling me is the power reduction formula gave me a sum of two cos functions and I wasn't sure what their combined frequency is, or period.  If there is a term of cos2theta, shouldn't the period be 1/2, since the frequency is doubled?  And if not, and the period is always 2pi, how do I express the recurring roots between the specified interval?
Thanks again for your help.  

    Wait, are you saying that because theta = pi/3, and the period is 2pi, then the following roots in the specified interval are 7pi/3 and 13pi/3?  Which means

I'm not the best at formalizing my math.    Is this correct?  

    Hmmmm, I think I'm missing some roots in the interval because I have a composite of a cos theta and a cos 2theta.  So should my answer beSomebody please help.  I'm having trouble seeing how the frequency of this equation can be ascertained so I can know where the roots are.  I used Wolfram Alpha to plot the equation, and I see the roots are 5pi/3, 7pi/3, 11pi/3, and 13pi/3, but I don't see how to get that.
Some one please help.  

    No, don't use the double angle  to find the answer, use the quadratic in  and do what you did earlier, you found  so now just find all the values of  that satisfy this equation in the domain specified.  

    Attached is the Wolfram plot  

                                   Originally Posted by Mentallic                   No, don't use the double angle  to find the answer, use the quadratic in  and do what you did earlier, you found  so now just find all the values of  that satisfy this equation in the domain specified.                  
But how do I find the values that count?  Wolphram also says but I don't know how to get that without looking at the plot.  How do I get this generalized term from the data given?

Thanks for your time.  

                                   Originally Posted by Gyro                   But how do I find the values that count?  Wolphram also says but I don't know how to get that without looking at the plot.  How do I get this generalized term from the data given?

Thanks for your time.                  
I think I get it now, maybe.
Since theta = pi/3, you check multiples of that in the equation, not multiples of 2pi like I thought.
So I plug in npi/3 for n = 1, 2, 3, 4, 5, ... and find all the values of n that make the equation 0 in the interval specified.  Then I have my roots as 5pi/3, 7pi/3, 11pi/3, and 13pi/3.  I can express each as
6pi/3 - pi/3, 6pi/3 + pi/3, 12pi/3 - pi/3, and 12pi/3 + pi/3.  If I factor out the 1/3 and use a plus/minus, I get to the general term I was looking for, 1/3(6npi +/- pi) for n = 1, 2 for the interval specified.  
Is this finally right?  

    Yes that looks good
You can also check the plot of  and see that it has the same roots as the original equation.  

    Thanks for your help.  

    It seems that this question is solved but, can't you just use the quadratic formula since you are using it on an interval?   

    Yep that's what Gyro has done, check post #9. And it doesn't matter if it's on an interval or not, you'll be using the quadratic formula regardless.
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