Find the points on the curve y=(cosx)/(2+sinx) at which the tangent is horizontal.
To find the points on the curve where the tangent is horizontal, we need to find the values of x where the derivative of y with respect to x (dy/dx) is equal to zero.
Let's start by finding the derivative of y with respect to x.
Given: y = (cosx)/(2+sinx)
To find dy/dx, we need to use the quotient rule for differentiation:
Quotient rule: d(u/v)/dx = (v * du/dx - u * dv/dx) / (v^2)
Let u = cosx and v = 2 + sinx.
dy/dx = [(2 + sinx)(-sinx) - (cosx)(cosx)] / (2 + sinx)^2
Now, equate dy/dx to zero and solve for x:
(2 + sinx)(-sinx) - (cosx)(cosx) = 0
Simplifying the equation:
-2sinx - sin^2(x) - cos^2(x) = 0
Using the trigonometric identity sin^2(x) + cos^2(x) = 1:
-2sinx - 1 = 0
-2sinx = 1
sinx = -1/2
We know that sin(pi/6) = 1/2, so we can conclude:
x = pi/6 + 2nπ or x = 5pi/6 + 2nπ, where n is an integer.
Now substitute these x-values back into the original equation y = (cosx)/(2+sinx) to find the y-coordinates:
When x = pi/6:
y = (cos(pi/6))/(2 + sin(pi/6))
= (√3/2)/(2 + 1/2)
= (√3/2) / (5/2)
= (√3/2) * (2/5)
= √3/5
When x = 5pi/6:
y = (cos(5pi/6))/(2 + sin(5pi/6))
= (-√3/2) / (2 - 1/2)
= (-√3/2) / (3/2)
= (-√3/2) * (2/3)
= -√3/3
Therefore, the points on the curve y = (cosx)/(2+sinx) where the tangent is horizontal are:
(pi/6, √3/5) and (5pi/6, -√3/3)