Question 1163480
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 3\cos\(2\theta\)\ +\ \cos\(\theta\)\ =\ 2]


Use the Double Angle Formula -- the form that only has cosine in it.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 3\(2\cos^2\(\theta\)\,-\,1\)\ +\ \cos\(\theta\)\ =\ 2]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 6\cos^2(\theta)\ +\ \cos(\theta)\ -\ 5\ =\ 0]


Let *[tex \Large u\ =\ \cos(\theta)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 6u^2\ +\ u\ -\ 5\ =\ 0]


The quadratic factors


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (u\ +\ 1)(6u\ -\ 5)\ =\ 0]


So


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \cos(\theta)\ =\ -1]


or


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \cos(\theta)\ =\ \frac{5}{6}]


And then


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \theta\ =\ \arccos(-1)\ =\ \pi]


or


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \theta\ =\ \arccos\(\frac{5}{6}\)]


or


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \theta\ =\ 2\pi\ -\ \arccos\(\frac{5}{6}\)]


These are the exact answers.  Use your calculator to find decimal approximations.  My answers are in radians; you can convert to degrees if you like.
								
								
John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
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