Quick Question..(Absolute Max/Min)

[Zerrotolerance]

I have a quick question. We are doing absolute max/min in calc class. I know how to find them, but am actually having trouble with the trig part. The equation is:

f(x) = x - 2 sin(x) on [0 , 2pi]
so,

f ' (x) = 1- 2cos(x)

I set this to 0 by EVT and solve to find the critical numbers.

I get to:

cos(x) = 1/2

Now, I don't know what to do to just find x. Can someone explain the process. When I put this in my calculator to solve I get some crazy output with symbols I have never seen before. (Something like: x = 6.28319*@n25+1.0472) This is probably basic, but it's driving me crazy. I know once I have the critical points I just plug them and the endpoints into f(x) to get absolute max and min. Any help on what I am supposed to do with this thanks!

 

[DrSteve]

\(\displaystyle cos(x) = 1/2\)

This is very basic trigonometry.

Draw a 30, 60, 90 triangle. The side opposite 30 is 1, the hypotenuse is 2, and the side opposite 60 is \(\displaystyle \sqrt{3}\).

Using cos x = opp/hyp, we see that 60 degrees satisfies the given equation.

But we need to convert this to radians...60 degrees = 60pi/180 = pi/3 radians.

There is another solution in the fourth quadrant (since cosine is positive in the fourth quadrant) of 2pi - pi/3 = 5pi/3.

So between 0 and 2pi we have the two solutions x = pi/3, x = 5pi/3.