Find x-Coordinate on (0,2π) for y=x√3+2sin(x) Horizontal Tangent Line

[cowgiljl]

y = x*square root of 3 + 2sinx find the x coordinate on (0,2pie) of the points where y has a horizonal tangent line

I am really lost and struggling on getting this started

thanks joe

 

[HallsofIvy]

Since y= x√(3)+ 2 sin x, y'= &radic(3)+ 3 cos x. The tangent line is horizontal when y'= 0.

 

[M98Ranger]

Hi Joe,

Don't worry, finding the x-coordinate for a horizontal tangent line can be a bit tricky at first. Let's break it down step by step.

First, we need to find the derivative of the given equation. The derivative will give us the slope of the tangent line at any given point.

So, let's start by finding the derivative of y = x√3+2sin(x):

y' = (√3 + 2cos(x)) * 1

We can simplify this to:

y' = √3 + 2cos(x)

Next, we need to find the x-values where the derivative is equal to 0. This is because when the derivative is 0, the slope of the tangent line is also 0, which means it is a horizontal line.

So, let's set the derivative equal to 0 and solve for x:

√3 + 2cos(x) = 0

2cos(x) = -√3

cos(x) = -√3/2

Now, we need to find the x-values on the interval (0,2π) that satisfy this equation. We can use a unit circle to find these values. The unit circle shows the cosine values for different angles, so we can use it to find the angles that have a cosine value of -√3/2.

Looking at the unit circle, we can see that the only angle on the interval (0,2π) that has a cosine value of -√3/2 is 5π/6.

So, the x-coordinate for the horizontal tangent line is x = 5π/6.

I hope this helps! Don't be discouraged, math can be challenging but with practice and patience, you can definitely master it. Keep up the good work!