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Naily [24]
2 years ago
9

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

Mathematics
1 answer:
Katen [24]2 years ago
3 0

Answer: 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.

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Glad I could help. Have A Good day

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Read 2 more answers
which expression represents purely real numbers and which expression represents non-real complex number?​
Lyrx [107]

Answer:

Step-by-step explanation:

Complex numbers are the numbers represented by the expression = a + bi

Here a = real part of the number

bi = imaginary part

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i^9=(i^2)^4.(i)

    =(-1)^{4}.i [Since, i^2=-1]

    =i

Therefore, its a non real complex number.

2 - \sqrt{-49} = 2-\sqrt{(-1)(7)^2}

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3 + 2i

It's a non real complex number.

-289

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