Question 828455: The question I'd like to ask is (5.3.2):
Solve the equation of 2cosx - 1 = 0
I know that I should have a set of two "x =" solutions, but need some help after that. I thought the answer might be "x = (pi/3) + 2*n*pi and x = (5pi/3)+2*n*pi (where n is an integer)", but am second guessing myself; I think it is close to that, but I might have a value or two off. If you could show me the answer, as well as how you go that answer, it would be much appreciated.
Answer by MathTherapy(10552)   (Show Source):
You can put this solution on YOUR website! The question I'd like to ask is (5.3.2):
Solve the equation of 2cosx - 1 = 0
I know that I should have a set of two "x =" solutions, but need some help after that. I thought the answer might be "x = (pi/3) + 2*n*pi and x = (5pi/3)+2*n*pi (where n is an integer)", but am second guessing myself; I think it is close to that, but I might have a value or two off. If you could show me the answer, as well as how you go that answer, it would be much appreciated.
2 cos x - 1 = 0
2 cos x = 1 ------ Adding 1 to both sides
1st Quadrant:  or 
4th Quadrant:  or 
You can do the check!!
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