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Math  /  Trigonometry

QuestionSolve the equation. Give a general formula for all the solutions. List six solutions. sin⁡θ=−12\sin \theta=-\frac{1}{2}
Identify the general formula for all the solutions to sin⁡θ=−12\sin \theta=-\frac{1}{2} based on the smaller angle. θ=\theta= □\square k is an integer (Simplify your answer. Use angle measureaz greater than or equal to 0 and less than 2π2 \pi. Type an exact answer, using π\pi as needed. Use integers or fractions for any numbers in the expression. Type an expression using kk as the variable.)

Studdy Solution

STEP 1

1. The equation sin⁡θ=−12\sin \theta = -\frac{1}{2} is a trigonometric equation.
2. We are looking for all solutions in terms of θ\theta over the interval [0,2π)[0, 2\pi).
3. The general solution will be expressed using kk, where kk is an integer.

STEP 2

1. Determine the reference angle.
2. Identify the quadrants where sin⁡θ=−12\sin \theta = -\frac{1}{2}.
3. Write the general formula for all solutions.
4. List six specific solutions.

STEP 3

First, determine the reference angle for sin⁡θ=12\sin \theta = \frac{1}{2}. The reference angle α\alpha is π6\frac{\pi}{6} because sin⁡π6=12\sin \frac{\pi}{6} = \frac{1}{2}.

STEP 4

Identify the quadrants where sin⁡θ\sin \theta is negative. sin⁡θ\sin \theta is negative in the third and fourth quadrants.

STEP 5

Using the reference angle π6\frac{\pi}{6}, find the angles in the third and fourth quadrants: - In the third quadrant, θ=π+π6=7π6\theta = \pi + \frac{\pi}{6} = \frac{7\pi}{6}. - In the fourth quadrant, θ=2π−π6=11π6\theta = 2\pi - \frac{\pi}{6} = \frac{11\pi}{6}.

STEP 6

Write the general formula for all solutions using the periodicity of the sine function, which has a period of 2π2\pi: - θ=7π6+2kπ\theta = \frac{7\pi}{6} + 2k\pi - θ=11π6+2kπ\theta = \frac{11\pi}{6} + 2k\pi

STEP 7

List six specific solutions by substituting integer values for kk into the general formulas:
1. For θ=7π6+2kπ\theta = \frac{7\pi}{6} + 2k\pi: - k=0k = 0: θ=7π6\theta = \frac{7\pi}{6} - k=1k = 1: θ=7π6+2π=19π6\theta = \frac{7\pi}{6} + 2\pi = \frac{19\pi}{6} - k=−1k = -1: θ=7π6−2π=−5π6\theta = \frac{7\pi}{6} - 2\pi = -\frac{5\pi}{6}
2. For θ=11π6+2kπ\theta = \frac{11\pi}{6} + 2k\pi: - k=0k = 0: θ=11π6\theta = \frac{11\pi}{6} - k=1k = 1: θ=11π6+2π=23π6\theta = \frac{11\pi}{6} + 2\pi = \frac{23\pi}{6} - k=−1k = -1: θ=11π6−2π=−π6\theta = \frac{11\pi}{6} - 2\pi = -\frac{\pi}{6}
The general formula for all solutions is: θ=7π6+2kπorθ=11π6+2kπ\theta = \frac{7\pi}{6} + 2k\pi \quad \text{or} \quad \theta = \frac{11\pi}{6} + 2k\pi
Six specific solutions are: 7π6,19π6,−5π6,11π6,23π6,−π6\frac{7\pi}{6}, \frac{19\pi}{6}, -\frac{5\pi}{6}, \frac{11\pi}{6}, \frac{23\pi}{6}, -\frac{\pi}{6}

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