Find the solutions for this equation in the interval [0, 2pi).
cos^3 x = cos x
Are they 0 and pi?
looks like it to me.
cos^3x - cosx = 0
cosx (cosx-1)(cosx+1) = 0
Oops. Looks like you left out where cosx=0.
0, pi/2, pi, and 3pi/2?
To solve the equation cos^3 x = cos x in the given interval [0, 2π), we can follow these steps:
1. Set cos^3 x - cos x = 0 since we want to solve the equation for x.
2. Rearrange the equation to obtain cos x (cos^2 x - 1) = 0.
3. Simplify cos^2 x - 1:
- Apply the trigonometric identity cos^2 x - sin^2 x = 1.
- Substitute 1 - sin^2 x for cos^2 x to get (1 - sin^2 x) - 1 = -sin^2 x.
4. Rewrite the equation as -sin^2 xcos x = 0.
5. Factor out -sin x: -sin x(sin x cos x) = 0.
Now, the equation can be true if either -sin x = 0 or sin xcos x = 0.
1. To solve -sin x = 0, we can find the values of x for which sin x = 0.
- In the interval [0, 2π), sin x = 0 at x = 0 and x = π.
2. To solve sin xcos x = 0, we need to find the values of x for which either sin x = 0 or cos x = 0.
- In the interval [0, 2π), cos x = 0 at x = π/2 and x = (3/2)π.
Combining all the values we found, the solutions for cos^3 x = cos x in the interval [0, 2π) are:
x = 0, x = π, x = π/2, and x = (3/2)π.
Therefore, the solutions are x = 0, x = π, x = π/2, and x = (3/2)π.