Question

At what points will `y = cos(2x) + sinx` have horizontal tangents?

Answer 1

`x = arcsin(1/4), pi/2, pi- arcsin(1/4), (3pi)/2`

Explanation:

Let's start by differentiating the function. Before doing this to the complete function, we need to find the derivative of the most complex part, `cos2x`.

Let `y = cosu` and `u = 2x`. `y' = -sinu` and `u' = 2`.

`dy/dx = 2 xx -sinu = -2sin(2x)`

We can now differentiate the entire function.

`y' = -2sin(2x) + (sinx)' = -2sin2x + cosx`

Now, we are looking for horizontal tangents. Horizontal tangents are horizontal because their slope is `0`, or the parameter `m` in `y = mx + b` equals `0`.

The slope of the tangent to a function at a given point `x = a` is given by evaluating `f(a)` into the derivative. We know the slope, so we can solve for `x` instead of solving for `y`.

`0 = -2sin2x + cosx`

Recall that `sin2x = 2sinxcosx`.

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

`0 = -4sinxcosx + cosx`

`0 = cosx(-4sinx + 1)`

`cosx = 0 and sinx = 1/4`

`x = pi/2, (3pi)/2, pi - arcsin(1/4) and arcsin(1/4)`

Hopefully this helps!