Finding real solutions to sin ⁡ x = x 50

Question

Finding real solutions to   sin  ⁡      x    =          x      50

Marlee Guerra · Accepted Answer

We can solve this problem using a Taylor series expansion to get approximate values:

\sin (x) = \frac{x}{50}

\sin (x) = \sum_{n = 1}^{\infty} \frac{(- 1)^{n + 1} x^{2 n - 1}}{(2 n - 1)!} = \frac{x}{1!} - \frac{x^{3}}{3!} + \frac{x^{5}}{5!} - \frac{x^{7}}{7!} + \dots

\frac{x}{1!} - \frac{x^{3}}{3!} + \frac{x^{5}}{5!} - \frac{x^{7}}{7!} + \dots = \frac{x}{50}

The trick here is to do approximations, not the actual thing.

\frac{x}{1!} = \frac{x}{50} \to x_{1} = 0

\frac{x}{1!} - \frac{x^{3}}{3!} = \frac{x}{50} \to x_{2} = 0, \pm \frac{7 \sqrt{3}}{5}

Repeating the process, we will get an infinite amount of answers, assuming we allow complex answers.
We will also note x=0 will always be a solution, one we can see graphically and within any of the above approximations we want.

Finding real solutions to sin ⁡<!-- ⁡

Answered question

Answer & Explanation

New Questions in Precalculus