This is my first time using this site, so if I'm not clear with the equations or formulas I present please let me know.
The problem is:
Find the value of x for F(x) =2cos(x)+1 where the tangent line is parallel to the secant line that passes through x=pi and x= 3pi/2, answers for x must me in the interval 0<=x<=2pi.
I started out plugging in x = pi and x= 3pi/2 gettng the points (pi, -1) and (3pi/2, 1)
using the slope formula y2-y1 \ x2-x1, I got the answer 4/pi for the slope of the secant line.
then I got the derivative using the constant = 0 rule to turn the 1 into a zero , the derivative of cos(x) = -sin(x) , and the constant multiplying a function rule to end up with F'(x) = -2sin(x)
Then I set 4/pi equal to -2sin(x) : 4/pi = -2sin(x) , 4/-2pi = sin(x) , -2/pi = sin(x)
and I'm stuck at this point.
I'm not sure how to isolate the x at this point. I'm pretty sure I have the work correct up to this point but I could be wrong, any help would be deeply appreciated.