how to solve $t(t−\sin t)^{5/3}=0.6\pi$

Question

How to solve this question? Using the Newton method? I have tried using this approximation, $\sin t ≈ 120(\pi−t)t/\pi^5$.

Answer 1

If you are looking for a numerical procedure, define $$ f(t) = t(t-\sin t)^{5/3} -2\pi/5 $$ and differentiate to find $f'(t)$. Then define the Newton recurrence relationship by $$ t_{n+1} = t_n - \frac{f(t_n)}{f'(t_n}, $$ plug in $f$ and $f'$ and simplify.

Can you pick a good initial guess $t_0$?

Answer 2

What, exactly, is the question? If it is just to solve $\theta\left(\theta- sin(\theta)\right)^{5/3}= 0$, the first thing I would do is to note that $\theta= 0$ is certainly a root. If $\theta \ne 0$ then we must have $\theta- sin(\theta)= 0$. To use "Newton's method", we have $f(\theta)= \theta- sin(\theta)$ so $f'(\theta)= 1- cos(\theta)$. Newton's method requires that we choose some starting value, $\theta_0$, then iteratively calculate $x_{n+1}- \frac{f(x_n)}{f'(x_n)}= \frac{\theta- sin(\theta)}{1- cos(\theta)}$. I don't see why you would use a quadratic approximation. In fact a graph quickly shows that x= 0 is the only root to the equation.