For the equation,
\[
2 \sin \theta=1
\]
a. Solve for all degree solutions.
b. Solve for $\theta$ if $0^{\circ} \leq \theta< 360^{\circ}$. Do not use a calculator.
Answer
Final Answer: The solutions to the equation \(2 \sin \theta=1\) for all degree solutions are \(\boxed{\theta = 30^\circ + 360n^\circ}\) and \(\boxed{\theta = 150^\circ + 360n^\circ}\), where \(n\) is an integer. The solutions for \(\theta\) in the range \(0^\circ \leq \theta<360^\circ\) are \(\boxed{\theta = 30^\circ}\) and \(\boxed{\theta = 150^\circ}\).
Step 1 ：First, we isolate \(\sin \theta\) by dividing both sides of the equation by 2. This gives us \(\sin \theta = \frac{1}{2}\).
Step 2 ：Next, we find the angle(s) whose sine is \(\frac{1}{2}\). We know that \(\sin 30^\circ = \frac{1}{2}\) and \(\sin 150^\circ = \frac{1}{2}\), so these are two possible solutions.
Step 3 ：However, the sine function is periodic with a period of \(360^\circ\), so we can add or subtract multiples of \(360^\circ\) to these solutions to get all possible degree solutions.
Step 4 ：For the second part of the question, we only need to consider the solutions in the range \(0^\circ \leq \theta<360^\circ\), which are \(30^\circ\) and \(150^\circ\).
Step 5 ：Final Answer: The solutions to the equation \(2 \sin \theta=1\) for all degree solutions are \(\boxed{\theta = 30^\circ + 360n^\circ}\) and \(\boxed{\theta = 150^\circ + 360n^\circ}\), where \(n\) is an integer. The solutions for \(\theta\) in the range \(0^\circ \leq \theta<360^\circ\) are \(\boxed{\theta = 30^\circ}\) and \(\boxed{\theta = 150^\circ}\).
