find the points(in exact form) at which the tangent is horizontal on the given interval:

Question

y=cos(2x)+2cos(x) , [0,2pi)

Answer 1

To find the points at which the tangent is horizontal, we need to find where the derivative of the function equals zero.

First, we find the derivative of the function:

y' = -2sin(2x) - 2sin(x)

Next, we set the derivative equal to zero and solve for x:

-2sin(2x) - 2sin(x) = 0

Factoring out a -2sin(x), we get:

-2sin(x)(cos(x) + cos(2x)) = 0

Setting each factor equal to zero, we have:

-2sin(x) = 0 cos(x) + cos(2x) = 0

From the first equation, sin(x) = 0. This occurs when x = 0, pi, and 2pi.

From the second equation, cos(x) + cos(2x) = 0. This equation is a bit more complicated to solve, but we can use the double-angle identities to simplify it. Using the identity cos(2x) = 2cos^2(x) - 1, we can rewrite the equation as:

cos(x) + 2cos^2(x) - 1 = 0

Rearranging the terms, we have:

2cos^2(x) + cos(x) - 1 = 0

Factoring the quadratic equation, we get:

(2cos(x) - 1)(cos(x) + 1) = 0

Setting each factor equal to zero, we have two possible solutions:

2cos(x) - 1 = 0 cos(x) + 1 = 0

Solving the first equation, we get:

2cos(x) = 1

cos(x) = 1/2

This occurs when x = pi/3 and x = 5pi/3.

Solving the second equation, we get:

cos(x) = -1

This occurs when x = pi.

Therefore, the points on the interval [0, 2pi) at which the tangent is horizontal are x = 0, x = pi/3, x = pi, and x = 5pi/3. To find the corresponding y-values, we substitute these x-values back into the original function:

For x = 0: y = cos(2(0)) + 2cos(0) = 1 + 2 = 3
For x = pi/3: y = cos(2(pi/3)) + 2cos(pi/3) = 1/2 + 1 = 3/2
For x = pi: y = cos(2(pi)) + 2cos(pi) = 1 + (-2) = -1
For x = 5pi/3: y = cos(2(5pi/3)) + 2cos(5pi/3) = 1/2 + 1 = 3/2

Therefore, the points at which the tangent is horizontal on the interval [0, 2pi) are (0, 3), (pi/3, 3/2), (pi, -1), and (5pi/3, 3/2).