For what values of x does the graph of f(x) = x - 2 sin x have a horizontal tangent?

Question

 (Use n as your integer variable. Enter your answers as a comma-separated list.) x=pi/3 +2pin, x=5pi/3 +2pin is not the right answer. I've tried many ways for the answer. 

Answer 1

f(x) = x - 2sinx

 

Horizontal tangent when f'(x) = 0

 

So, we need to solve the equation 1 - 2cosx = 0

 

                                                       cosx = 1/2

 

Solutions in the interval [0, 2π] are π/3, 5π/3

 

All solutions:  π/3 + 2nπ, 5π/3 + 2nπ, where n = any integer

 

 

Answer 2

You will take the derivative of f(x) equal to zero.  This is because derivative is the slope of the tangent line, and horizontal lines have a slope of 0.
 
f(x) = x - 2sin(x)
 
 
f'(x) = 0
 
1 - 2cos(x) = 0
 
 
No just solve for x from this bolded equation.
 
-2cos(x) = -1
 
  cos(x) = 1/2
 
        x = π/3
 
The values of x are also known as your critical points.  They represent maximum and minimum values.
 
 
Cosine is positive in the 1st and 4th quadrants.
 
x = 2π - (π/3) = (6π - π) / 3 = 5π/3
 
x = 4π - (π/3) = (12π - π) / 3 = 11π/3
 
x = 6π - (π/3) = (18π - π) / 3 = 17π/3
 
 
So the value of x will be
 
x = (π/3) + nπ    ---->  1st quadrant

 

where   n is all even integers greater than or equal to zero.

 

 

x = nπ - (π/3)     ----->  4th quadrant
 
where   n is all even integers greater than zero.

 

 

x = nπ - (π/3)     ----->  4th quadrant