Some number belongs to real such that

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SUMMARY

The discussion centers on the equation sin(x) = x - 1, where the user analyzes the intervals of sin(x) and x - 1 within the range [0, 2π]. The user correctly identifies that sin(x) is bounded between -1 and 1, leading to the conclusion that there exists a real number x within the interval (-1, 1). However, the assertion that no such x exists between -1 and 1 is incorrect; the existence of a solution is confirmed through graphical analysis of y = sin(x) and y = x - 1.

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  • Understanding of trigonometric functions, specifically sine.
  • Knowledge of interval notation and real number properties.
  • Familiarity with graphical analysis of functions.
  • Basic algebraic manipulation skills.
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  • Explore the graphical intersection of y = sin(x) and y = x - 1 using graphing software.
  • Learn about the Intermediate Value Theorem and its application in finding roots of equations.
  • Study numerical methods for approximating solutions to equations, such as the Newton-Raphson method.
  • Investigate the properties of periodic functions and their intersections with linear functions.
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Homework Statement



sin(x) = x - 1

Homework Equations





The Attempt at a Solution



i used the fact that -1 < sin(x) < 1
and set interval to [0, 2pi].

this gave me,

-1 < sin(x) < 1 and -1 < x - 1 < 2pi - 1
so therefore, since sin(x) < 1 < 2pi - 1
there must be x, real number that lies between -1 and 1.


is this correct?
 
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Draw a graph of y=sin(x) versus y=x-1.
 
You are correct in proving that such a x exists. Do you also need to find that x? That seems a bit difficult...
 
There is no such x between -1 and 1. That is not to say that such an x does not exist.
 

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