For what values of x∈[0,2π] does the graph of y=x+2cosx have a horizontal tangent?

x = pi/6 and x = pi/3.

Step-by-step explanation:

We have the function:

y = x + 2*cos(x)

It will have a horizontal tangent at the point where it's derivate is equal to zero.

Then first let's differentiate y.

y' = dy/dx = 1 - 2*sin(x).

Then we must find the value of x between 0 and 2*pi (or 0° and 360°)

y'(x) = 1 - 2*sin(x) = 0.

Let's solve that:

2*sin(x) = 1

sin(x) = 1/2.

We know that:

sin(30°) = 1/2.

and

Sin(120°) = 1/2

Then let's convert 30° into radians.

We know that:

pi = 180°.

Then:

pi/180° = 1.

30° = 30°*(pi/180°) = (30°/180°)*pi = (3/18)*pi = pi/6

120° = (120°/180°)*pi = (12/18)*pi = (1/3)*pi = pi/3.

Then the two values of x are: pi/6 and pi/3.

fa tha first one you gonna add y to both sides

it look like this

x-y+y=6+y

x=y+6

therefore your answer is

x =y+6

fa tha second one

you gonna flip tha equation

it gonna look like x-4y=y

then add 4 to both sides

it gonna look like tha

x-4 +4 =y +4

so therefore tha answer is x=y+4

step-by-step explanation:

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