Find the general solution for $ heta$ in the equation $sin( heta) = frac{1}{2}$ using the unit circle.

To find the general solution for the equation $\sin(\theta) = \frac{1}{2}$, we use the unit circle. On the unit circle, $\sin(\theta) = \frac{1}{2}$ at $\theta = \frac{\pi}{6}$ and $\theta = \frac{5\pi}{6}$ within one period $[0, 2\pi)$.

Therefore, the general solutions are:

$\theta = \frac{\pi}{6} + 2k\pi, \quad k \in \mathbb{Z}$

and

$\theta = \frac{5\pi}{6} + 2k\pi, \quad k \in \mathbb{Z}$

To solve $sin( heta) = frac{1}{2}$, we refer to the unit circle. We know that $sin( heta)$ reaches $frac{1}{2}$ at $ heta = frac{pi}{6}$ and $ heta = frac{5pi}{6}$ within the interval $[0, 2pi)$.

Thus, the general solutions are:

$ heta = frac{pi}{6} + 2npi, quad n in mathbb{Z}$

and

$ heta = frac{5pi}{6} + 2npi, quad n in mathbb{Z}$

The equation $sin( heta) = frac{1}{2}$ is satisfied at $ heta = frac{pi}{6}$ and $ heta = frac{5pi}{6}$ in the interval $[0, 2pi)$.

The general solutions are:

$ heta = frac{pi}{6} + 2mpi, quad m in mathbb{Z}$

and

$ heta = frac{5pi}{6} + 2mpi, quad m in mathbb{Z}$

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