To solve the equation \( \sin(2\theta) = -\frac{1}{2} \), we can use the double angle identity for sine: \( \sin(2\theta) = 2\sin(\theta)\cos(\theta) \).
So, we have \( 2\sin(\theta)\cos(\theta) = -\frac{1}{2} \).
Since \( \sin \) is negative in the second and third quadrants, and we are given that \( \sin(2\theta) = -\frac{1}{2} \), we know that \( 2\theta \) lies in these quadrants. This means that \( \theta \) must lie in the second and third quadrants.
We can use the unit circle or trigonometric identities to solve for \( \theta \). Taking the inverse sine of \( -\frac{1}{2} \) gives us \( \frac{7\pi}{6} \) and \( \frac{11\pi}{6} \).
Therefore, the general formula for the solutions is \( \theta = \frac{7\pi}{12} + \frac{n\pi}{2} \) and \( \theta = \frac{11\pi}{12} + \frac{n\pi}{2} \), where \( n \) is an integer.
Six solutions within the interval \( [0, 2\pi) \) are:
1. \( \theta = \frac{7\pi}{12} \)
2. \( \theta = \frac{11\pi}{12} \)
3. \( \theta = \frac{19\pi}{12} \)
4. \( \theta = \frac{23\pi}{12} \)
5. \( \theta = \frac{31\pi}{12} \)
6. \( \theta = \frac{35\pi}{12} \)
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Part 1
Solve the equation. Give a general formula for all the solutions. List six solutions.
Sin\left(2\theta \right)=-\frac{1}{2}
1 answer
1 answer