SOLUTION: Find all values of x to the nearest degree that satisfy the equation 3cos2x + cosx + 2 = 0 if 0o < x

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Question 613354: Find all values of x to the nearest degree that satisfy the equation 3cos2x + cosx + 2 = 0 if 0o < x < 360o.
Found 3 solutions by htmentor, solver91311, Alan3354:
Answer by htmentor(1343)   (Show Source):
You can put this solution on YOUR website!
Find all values of x to the nearest degree that satisfy the equation 3cos2x + cosx + 2 = 0 if 0o < x < 360
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Using the identity cos(2x) = 2cos^2(x) - 1, we can write the equation as
3(2cos^2(x)-1) + cos(x) + 2 = 0
Simplifying gives
6cos^2(x) + cos(x) - 1 = 0
This can be factored as
(3cos(x)-1)(2cos(x)+1) = 0
This gives cos(x) = 1/3 and cos(x) = -1/2
The 1st 2 angles are 71 and 120 deg.
I'll leave it as an exercise to figure out the rest (there are 2 more).

Answer by solver91311(24713)   (Show Source):
You can put this solution on YOUR website!

Depends. Do you mean

or

John

My calculator said it, I believe it, that settles it

Answer by Alan3354(69443)   (Show Source):
You can put this solution on YOUR website!
Find all values of x to the nearest degree that satisfy the equation 3cos2x + cosx + 2 = 0 if 0 < x < 360
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If it's and not 3cos(2x),
Sub u for cos(x)
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cos(x) = u = complex numbers
I doubt you want the arccos of complex numbers.
Clarify the first term with parentheses.

SOLUTION: Find all values of x to the nearest degree that satisfy the equation 3cos2x + cosx + 2 = 0 if 0o < x < 360o.