Quadratic equation to find the roots of cos theta

[Gyro]

Homework Statement

Solve the equation $$2cos^2\theta + 5cos\theta - 3 = 0\ for\ \frac{3\pi}{2}<\theta<5\pi$$

Homework Equations

quadratic equation,
power reduction formula:
$$cos^2\theta = \frac {1+cos2\theta}{2}$$

The Attempt at a Solution

First I tried using the quadratic equation to find the roots of cos theta and got:
$$cos\theta = 1/2, \theta = \pi /3$$ 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 $$cos2\theta +5cos\theta = 2$$
but I don't know where to go from there. Can anyone help me please?

 

[Mentallic]

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.

 

[Gyro]

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.

 

[Gyro]

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
$$\theta = \frac {2n\pi + \pi} {3} for\ n = 3 ,\ 6?$$
I'm not the best at formalizing my math. Is this correct?

 

[Gyro]

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 be
$$\theta = \frac {2n\pi + \pi}{3}\ for\ n = 3, 4, 5, 6, 7, \ and\ 8$$

Somebody 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.

 

[Mentallic]

No, don't use the double angle $\cos2\theta$ to find the answer, use the quadratic in $\cos\theta$ and do what you did earlier, you found $\cos\theta=1/2$ so now just find all the values of $\theta$ that satisfy this equation in the domain specified.

 

[Gyro]

Attached is the Wolfram plot

 

[Gyro]

No, don't use the double angle $\cos2\theta$ to find the answer, use the quadratic in $\cos\theta$ and do what you did earlier, you found $\cos\theta=1/2$ so now just find all the values of $\theta$ that satisfy this equation in the domain specified.

But how do I find the values that count? Wolphram also says
$$\theta = \frac {1}{3}(6n\pi \pm \pi)$$ 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.

 

[Gyro]

But how do I find the values that count? Wolphram also says
$$\theta = \frac {1}{3}(6n\pi \pm \pi)$$ 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?

 

[Mentallic]

Yes that looks good
You can also check the plot of $$y=\cos(x)-1/2$$ and see that it has the same roots as the original equation.

 

[Gyro]

Thanks for your help.

 

[MysticDude]

It seems that this question is solved but, can't you just use the quadratic formula since you are using it on an interval? $$\frac{-5 +- \sqrt{25 +24}}{4}$$

 

[Mentallic]

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.