Question

How do you find the x-coordinates of all points on the curve `y=sin2x-2sinx` at which the tangent line is horizontal?

Answer 1

In the interval `0 ≤ x ≤ 2pi`, the points are `x= (5pi)/6, (7pi)/6 and 0`.

Explanation:

The slope of the tangent where it is horizontal is `0`. Since the output of the derivative is the instantaneous rate of change (slope) of the function, we will need to find the derivative.

`y = sin2x - 2sinx`

Let `y = sinu` and `u = 2x`. By the chain rule, `dy/dx= cosu xx 2 = 2cos(2x)`.

`y' = 2cos(2x) - 2cosx`

We now set `y'` to `0` and solve for `x`.

`0 = 2cos(2x) - 2cosx`

Apply the identity `cos(2beta) = 2cos^2beta - 1`

`0 = 2(2cos^2x - 1) - 2cosx`

`0 = 4cos^2x - 2cosx - 2`

`0 = 2(2cos^2x - cosx - 1)`

`0 = 2cos^2x - 2cosx + cosx - 1`

`0 = 2cosx(cosx- 1) + 1(cosx - 1)`

`0= (2cosx + 1)(cosx - 1)`

`cosx= -1/2 and cosx = 1`

`x = (5pi)/6, (7pi)/6 and 0`

Hopefully this helps!