SOLUTION: Solve 3cos2𜃠+ cos𜃠= 2 for 0 < 𜃠< 360°.
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Question 1163480: Solve 3cos2𜃠+ cos𜃠= 2 for 0 < 𜃠< 360°.
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
\ +\ \cos\(\theta\)\ =\ 2)
Use the Double Angle Formula -- the form that only has cosine in it.
\,-\,1\)\ +\ \cos\(\theta\)\ =\ 2)
\ +\ \cos(\theta)\ -\ 5\ =\ 0)
Let )
The quadratic factors
(6u\ -\ 5)\ =\ 0)
So
\ =\ -1)
or
\ =\ \frac{5}{6})
And then
\ =\ \pi)
or
)
or
)
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
My calculator said it, I believe it, that settles it
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