$Find all angles heta in radians on the unit circle where sin( heta) = frac{1}{2} and cos( heta) = -frac{sqrt{3}}{2}$

To find angles θ where $\sin(\theta) = \frac{1}{2}$ and $\cos(\theta) = -\frac{\sqrt{3}}{2}$, we start by identifying possible angles for each trigonometric condition separately:

From $\sin(\theta) = \frac{1}{2}$, the possible angles are $\theta = \frac{\pi}{6}$ and $\theta = \frac{5\pi}{6}$ in the first and second quadrants.

From $\cos(\theta) = -\frac{\sqrt{3}}{2}$, the possible angles are $\theta = \frac{5\pi}{6}$ and $\theta = \frac{7\pi}{6}$ in the second and third quadrants.

The common angle satisfying both conditions is $\theta = \frac{5\pi}{6}$. Therefore, the solution is:

$ \boxed{\frac{5\pi}{6}} $

First, let’s identify the angles that satisfy $sin( heta) = frac{1}{2}$. These angles, on the unit circle, occur at:

$ heta = frac{pi}{6} or heta = frac{5pi}{6}$

Next, we identify the angles that satisfy $cos( heta) = -frac{sqrt{3}}{2}$. These angles occur at:

$ heta = frac{5pi}{6} or heta = frac{7pi}{6}$

The angle that satisfies both $sin( heta) = frac{1}{2}$ and $cos( heta) = -frac{sqrt{3}}{2}$ is:

$ oxed{frac{5pi}{6}} $

Combining the conditions $sin( heta) = frac{1}{2}$ and $cos( heta) = -frac{sqrt{3}}{2}$, we find the solution:

$ oxed{frac{5pi}{6}} $

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