Find all solutions to the equation in the interval [0,2pi)
Cosx-cos2x
I will assume your "equation" is
cosx - cos 2x = 0
cosx - (2cos^2 x - 1) = 0
2cos^2 x - cosx - 1 = 0
(2cosx + 1)(cosx - 1) = 0
cosx = -1/2 or cosx = 1
if cosx = -1/2 , then
x = 2π/3 or x = 4π/3 , (120° or 240°)
if cosx = 1 , then
x = 0 or 2π , (0° or 360°)
Thanks!
To find all the solutions to the equation cos(x) - cos(2x) = 0 in the interval [0, 2π), we need to determine the values of x that satisfy the equation.
Step 1: Use the trigonometric identity cos(2x) = 2cos^2(x) - 1.
Rewrite the equation as: cos(x) - (2cos^2(x) - 1) = 0.
Simplify the equation to: 2cos^2(x) - cos(x) - 1 = 0.
Step 2: Solve the quadratic equation.
Let's consider the quadratic equation 2cos^2(x) - cos(x) - 1 = 0.
This equation can be factored as: (2cos(x) + 1)(cos(x) - 1) = 0.
Set each factor equal to zero:
2cos(x) + 1 = 0 => cos(x) = -1/2
cos(x) - 1 = 0 => cos(x) = 1
Step 3: Solve for x in each equation.
To find the solutions for cos(x) = -1/2, we need to determine where the cosine function takes that value in the given interval [0, 2π). We can use the unit circle or cosine graph to identify the solutions. The solutions are x = π/3 and x = 5π/3.
To find the solutions for cos(x) = 1, we need to identify where the cosine function takes that value in the interval [0, 2π). The solutions are x = 0 and x = 2π.
Step 4: Combine all the solutions.
The solutions in the interval [0, 2π) to the equation cos(x) - cos(2x) = 0 are x = 0, x = π/3, x = 5π/3, and x = 2π.
Therefore, the solutions to the equation cos(x) - cos(2x) = 0 in the interval [0, 2π) are x = 0, x = π/3, x = 5π/3, and x = 2π.