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.