Question

Find the points at which the tangent is horizontal? y = (cos(x))/(2 + sin(x))

Answer 1

Considering trig functions are periodic, the tangents will be horizontal at

`x = pi/6 + 2pin`, n an integer
`x = (5pi)/6 + 2pin`, n an integer

Explanation:

We start by finding the derivative using the quotient rule.

`y' = (-sinx(2 + sinx) - cosx(cosx))/(2 + sinx)^2`

`y' = (-2sinx - sin^2x - cos^2x)/(2 + sinx)^2`

`y' = (-(2sinx + sin^2x + cos^2x))/(2 + sinx)^2`

`y' = - (1 - 2sinx)/(2 + sinx)^2`

Recall that the derivative represents the instantaneous rate of change of the function at any given point in its domain. Also recall that a horizontal line (e.g. `y = a`), has slope `0` (since its equation can be written `y = 0x + a`, and this is the form `y = mx + b`, where m is the slope).

Set the derivative to `0` and solve for `x`.

`0 = - (1 - 2sinx)/(2 + sinx)^2`

`0(2 + sinx)^2 = -(1 - 2sinx)`

`0 = 2sinx - 1`

`sinx = 1/2`

`x = pi/6 + 2pin, (5pi)/6 + 2pin`

Hopefully this helps!