Question

The number of solutions of 2
of 200s? sinx=x?
infinite​

Answer 1

answer:

If the problem could be solved by purely algebraic means (with a finite number of steps), that would imply that sin(x)sin⁡(x) could be given a polynomial representation from which you could go about your usual routine of factoring to find the zeroes of the polynomial.

The interesting point here is that no such representation for sin(x)sin⁡(x) exists, unless you are okay with it being infinitely long.

The trigonometric functions like sin()sin⁡() and cos()cos⁡() are part of a category of transcendental functions--so called because they transcend the expressive power of algebra to describe them.

step-by-step explaination:

Here's a shot at solving it algebraically if we can cheat and use a result from calculus:

Given this identity:

sin(x)=x−x33!+x55!−x77!+⋯

sin⁡(x)=x−x33!+x55!−x77!+⋯

Subtract out your problem sin(x)=xsin⁡(x)=x

0=−x33!+x55!−x77!+⋯

0=−x33!+x55!−x77!+⋯

0=x3(−13!+x25!−x47!+⋯)

0=x3(−13!+x25!−x47!+⋯)

x3=0or(−13!+x25!−x47!+⋯)=0

x3=0or(−13!+x25!−x47!+⋯)=0

So now we have our "algebraic solution" that x=0x=0.