cos2x=cosx in interval of [0,2pie]

cos2x-cosx=0

expand the left hand side
cos²(x)-sin²(x)-cos(x)=0
2cos²(x)-1 - cos(x)=0
substitute c=cos(x)
2c²-c-1=0
c=(-1±√(9))/4
=-1 or 1/2
cos(x)=-1 when x=π (0≤x≤2π)
or
cos(x)=1/2 when x=π/3 or x=5π/3 (0≤x≤2π)

Substitute each of the three solution into the original equation to make sure that the solutions are acceptable.

Reiny is right.

There was a mistake in the solution of the quadratic.

c=(1±√(9))/4
=1 or -1/2
cos(x)=1 when x=0 or 2π (0≤x≤2π)
or
cos(x)=-1/2 when x=π±π/3 (0≤x≤2π)

Substitute each of the three solution into the original equation to make sure that the solutions are acceptable.

To find the values of x in the interval [0, 2π] that satisfy the equation cos(2x) = cos(x), we can use trigonometric identities and solve for x.

First, let's simplify the given equation using the identity cos(2x) = 2cos^2(x) - 1:

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

Now, rearrange the equation to form a quadratic equation:

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

Let's substitute cos(x) with a variable, let's say u:

2u^2 - u - 1 = 0

Now, solve this quadratic equation for u. We can either factorize it or use the quadratic formula.

Let's use the quadratic formula: u = (-b ± √(b^2 - 4ac))/(2a)

For our equation, a = 2, b = -1, and c = -1.

u = (-(-1) ± √((-1)^2 - 4(2)(-1)))/(2(2))
= (1 ± √(1 + 8))/4
= (1 ± √9)/4
= (1 ± 3)/4

We have two possible solutions for u:

1) u = (1 + 3)/4 = 4/4 = 1
2) u = (1 - 3)/4 = -2/4 = -1/2

Now we substitute back u with cos(x) since u represents the values of cos(x):

1) cos(x) = 1
2) cos(x) = -1/2

To find the values of x, we need to take the inverse cosine (or arccos) of each equation.

1) x = arccos(1) = 0
2) x = arccos(-1/2)

To find the values of x for the second equation, we can use the unit circle or evaluate it numerically using a calculator.

Using the unit circle, we can see that arccos(-1/2) has two solutions: x = 2π/3 and x = 4π/3.

Therefore, the solutions to cos(2x) = cos(x) in the interval [0, 2π] are:

x = 0, 2π/3, and 4π/3.