t=sin(2*t^3)
get the t
Wow, messy
I used
http://www.math.hawaii.edu/lab/241/online-grapher.shtml
and the function
sin(2*x^3) - x
to get answers of x=0 and
x= appr. 0.75 and x = appr 0.97
I further fine-tuned the last two by trial and error and got
x = 0.7523 to get
LS = .7523
RS = .75229 , so that answer is correct to 4 decimals
I will let you do the same for the last approximation
Btw, because of the symmetry of the function
x = appr -0.75 and x = appr. -0.97 would also work
To solve the equation t = sin(2*t^3) and find the value of t, we need to use numerical methods. One common numerical root-finding method is the Newton-Raphson method.
Here are the steps to apply the Newton-Raphson method:
1. Start with an initial guess for the value of t. Let's call this initial guess t_0.
2. Calculate the function value at this initial guess using t_0. Let's call this f(t_0).
f(t_0) = sin(2*t_0^3)
3. Calculate the derivative of the function at t_0. Let's call this f'(t_0).
f'(t_0) = d/dt [sin(2*t^3)]
4. Update the guess for t using the Newton-Raphson iteration formula:
t_1 = t_0 - (f(t_0) / f'(t_0))
5. Repeat steps 2-4 until you reach a desired level of accuracy.
In this case, since the equation is transcendental (cannot be solved algebraically), we will need to use iterative numerical methods to approximate the value of t. The Newton-Raphson method is a common choice for root-finding problems like this.
Note that the Newton-Raphson method may not always guarantee convergence or find all solutions, depending on the initial guess and the nature of the function. It may be necessary to try different initial guesses to find all possible solutions.
Also, keep in mind that the equation t = sin(2*t^3) is a transcendental equation and may have multiple solutions or no solutions depending on the desired range or constraints.