Math Problem Statement

if sin x = -1 and x=[0,2pi] what is the value of x

Solution

The equation sinx=1\sin x = -1 within the interval [0,2π][0, 2\pi] asks for the values of xx where the sine function equals -1.

The sine function reaches a value of -1 at the angle x=3π2x = \frac{3\pi}{2} (which is 270° in degrees).

Thus, the value of xx in the interval [0,2π][0, 2\pi] is:

x=3π2x = \frac{3\pi}{2}

Would you like more details or have any questions? Here are 5 related questions to explore further:

  1. What is the general solution to sinx=1\sin x = -1 for all possible values of xx?
  2. How can the unit circle help determine the sine values for specific angles?
  3. What are the properties of the sine function in different quadrants?
  4. How can we solve sinx=k\sin x = k for arbitrary values of kk?
  5. What is the relationship between the sine and cosine functions?

Tip: The sine function has a period of 2π2\pi, meaning it repeats every 2π2\pi units.

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Math Problem Analysis

Mathematical Concepts

Trigonometry
Sine Function
Unit Circle

Formulas

sin(x) = -1

Theorems

Sine Function Periodicity

Suitable Grade Level

Grades 10-12