Math Problem Statement

Using the graph of y = sin(x), list all values on the interval [11π/6, 13π/4] that satisfy the ordered pair (x₁, -1/2).

Solution

The problem asks for values of x1x_1 where sin(x1)=12\sin(x_1) = -\frac{1}{2} within the interval [11π6,13π4]\left[ \frac{11\pi}{6}, \frac{13\pi}{4} \right].

Step 1: Solve sin(x1)=12\sin(x_1) = -\frac{1}{2}

The general solution for sin(x1)=12\sin(x_1) = -\frac{1}{2} is: x1=2kπ+7π6orx1=2kπ+11π6x_1 = 2k\pi + \frac{7\pi}{6} \quad \text{or} \quad x_1 = 2k\pi + \frac{11\pi}{6} for any integer kk, since these angles correspond to where sin(x)=12\sin(x) = -\frac{1}{2}.

Step 2: Find values of x1x_1 in the interval [11π6,13π4]\left[ \frac{11\pi}{6}, \frac{13\pi}{4} \right]

We need to convert the upper bound 13π4\frac{13\pi}{4} to a more comparable form: 13π4=3π+π4=12π4+π4=13π4\frac{13\pi}{4} = 3\pi + \frac{\pi}{4} = \frac{12\pi}{4} + \frac{\pi}{4} = \frac{13\pi}{4} Thus, the interval is approximately [5.76,10.2][5.76, 10.2] (in radians).

Step 3: Determine specific values

By adding or subtracting multiples of 2π2\pi, we will determine which solutions fall into the given interval.

Let me calculate this for you.The values of x1x_1 that satisfy sin(x1)=12\sin(x_1) = -\frac{1}{2} within the interval [11π6,13π4]\left[ \frac{11\pi}{6}, \frac{13\pi}{4} \right] are approximately:

x15.76andx19.95x_1 \approx 5.76 \quad \text{and} \quad x_1 \approx 9.95

These correspond to angles where the sine function equals 12-\frac{1}{2} in the given interval.

Would you like further details or have any questions?

Here are 5 related questions to expand on the topic:

  1. What is the general solution for the equation sin(x)=1\sin(x) = -1?
  2. How do we determine the period of trigonometric functions like sine and cosine?
  3. What is the reference angle for the sine function in the unit circle?
  4. How do we solve cos(x)=12\cos(x) = \frac{1}{2} within a specified interval?
  5. Can you explain how to convert radians to degrees and vice versa?

Tip: When solving trigonometric equations over intervals, remember to adjust for the periodic nature of the function by considering all solutions that fall within the interval.

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Math Problem Analysis

Mathematical Concepts

Trigonometry
Sine Function
Unit Circle
Solving Trigonometric Equations

Formulas

sin(x) = -1/2
General solution for sine: x = nπ + (-1)^n * arcsin(value)

Theorems

Periodic nature of sine function
General solution of sine equations

Suitable Grade Level

Grades 9-12