\[
2 \sin \theta+1=\csc \theta
\]
Select the correct choice below and, if necessary, fill in the answer box to complete your choice.
A. The solution set is $\{\square$.
(Type your answer in degrees. Do not include the degree symbol in your answer. Round to one decimal place as needed. Use a comma to separate answers as needed.)
B. The solution is the empty set

Step 1 ：Let's denote \(a = \sin \theta\) and \(b = \cos \theta\), so the equation becomes \(2a + 1 = \frac{1}{a}\).

Step 2 ：Rearranging the terms, we get \(2a^2 - a + 1 = 0\).

Step 3 ：This is a quadratic equation in terms of \(a\), which can be solved using the quadratic formula \(a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). Here, \(a = 2\), \(b = -1\), and \(c = 1\).

Step 4 ：Substituting these values into the quadratic formula, we get \(a = \frac{1 \pm \sqrt{1 - 4(2)(1)}}{2(2)} = \frac{1 \pm \sqrt{-7}}{4}\).

Step 5 ：Since the square root of a negative number is not a real number, there are no real solutions for \(a\).

Step 6 ：Since \(a = \sin \theta\), and \(\sin \theta\) must be a real number, there are no solutions for \(\theta\) that satisfy the original equation.

Step 7 ：So, the solution is the empty set.