Find all solutions to the equation
2cos(2x) = x + 1, correct to 4 decimal places.
2 cos(2x) never gets out of the range [-2,2]
So, whenever |x+1| > 2, there are no solutions.
That means all solutions are in [-3,1]
A little Newton-Raphson should work nicely here, or a bisection method.
To find all solutions to the equation 2cos(2x) = x + 1, we'll have to use numerical methods since this equation cannot be solved analytically. One numerical method that can be used is the Newton-Raphson Method.
The Newton-Raphson Method is an iterative numerical method that can find solutions to equations by repeatedly applying a formula. Here's how we can use this method to find the solutions to the given equation:
1. Start with an initial guess for the solution, let's say x₀ = 0.
2. Iterate using the formula xᵢ₊₁ = xᵢ - f(xᵢ)/f'(xᵢ), until we reach a desired level of accuracy.
- In our case, the equation is 2cos(2x) = x + 1. So, f(x) = 2cos(2x) - (x + 1).
- To find f'(x), we need to differentiate f(x) = 2cos(2x) - (x + 1) with respect to x. The derivative is f'(x) = -4sin(2x) - 1.
- Substitute these values into the Newton-Raphson formula, xᵢ₊₁ = xᵢ - (2cos(2xᵢ) - (xᵢ + 1))/(-4sin(2xᵢ) - 1).
3. Repeat step 2 until the difference between successive approximations is smaller than the desired level of accuracy. In this case, we want the solutions correct to 4 decimal places.
By following these steps, we can apply the Newton-Raphson Method using numerical software (e.g., Python or Mathematica) to find the approximations of the solutions to the given equation.