the number As cond degree trigonometric equation is written in the form $\mathrm{n}^{2} \theta+B \sin \theta+C=0$, where $A>0$. When the equation is factored, one of the fät $\mathrm{rs}$ is $(\sin \theta+2)$. The equation has only two roots, $\theta=\frac{7 \pi}{6}, \frac{11 \pi}{6}$, on the don ain of $0<\theta<2 \pi$. Numeric Response 1. Determine the values of $A, B$ and $C$ for the equation of the second degree trigonometric equation in the form $A \sin ^{2} \theta+B \sin \theta+C=0$. The values of $A, B$, and $C$, respectively, are (Record your answer in the numerical-response section below.) Your answer:

Step 1 ：Given that one of the factors of the equation is \((\sin \theta + 2)\), we can write the equation as \(A(\sin \theta + 2)(\sin \theta + k) = 0\), where \(k\) is the other factor.

Step 2 ：Since the equation has roots \(\theta = \frac{7\pi}{6}, \frac{11\pi}{6}\), we substitute these values into the equation to find \(k\).

Step 3 ：For \(\theta = \frac{7\pi}{6}\), we have \(A(\sin \frac{7\pi}{6} + 2)(\sin \frac{7\pi}{6} + k) = 0\), which simplifies to \(A(-\frac{1}{2} + 2)(-\frac{1}{2} + k) = 0\). This gives us \(A(\frac{3}{2})(\frac{1}{2} + k) = 0\).

Step 4 ：For \(\theta = \frac{11\pi}{6}\), we have \(A(\sin \frac{11\pi}{6} + 2)(\sin \frac{11\pi}{6} + k) = 0\), which simplifies to \(A(\frac{1}{2} + 2)(\frac{1}{2} + k) = 0\). This gives us \(A(\frac{5}{2})(-\frac{1}{2} + k) = 0\).

Step 5 ：Setting these two equations equal to each other, we get \(A(\frac{3}{2})(\frac{1}{2} + k) = A(\frac{5}{2})(-\frac{1}{2} + k)\). Solving for \(k\), we find that \(k = -2\).

Step 6 ：Substituting \(k = -2\) back into the equation, we get \(A(\sin \theta + 2)(\sin \theta - 2) = 0\), which simplifies to \(A\sin^2 \theta - 4A = 0\).

Step 7 ：Comparing this with the original form of the equation \(A\sin^2 \theta + B\sin \theta + C = 0\), we can see that \(B = 0\) and \(C = -4A\).

Step 8 ：Since the equation has only two roots, we know that the discriminant \(B^2 - 4AC = 0\). Substituting \(B = 0\) and \(C = -4A\) into this equation, we get \(0 - 4A(-4A) = 0\), which simplifies to \(16A^2 = 0\). Solving for \(A\), we find that \(A = 0\).

Step 9 ：However, we are given that \(A > 0\), so there must be a mistake. Looking back at our steps, we realize that we made a mistake in step 6. The equation should be \(A\sin^2 \theta - 4A\sin \theta + 4A = 0\), not \(A\sin^2 \theta - 4A = 0\).

Step 10 ：Correcting this, we see that \(B = -4A\) and \(C = 4A\).

Step 11 ：Substituting \(B = -4A\) and \(C = 4A\) into the discriminant equation \(B^2 - 4AC = 0\), we get \((-4A)^2 - 4A(4A) = 0\), which simplifies to \(16A^2 - 16A^2 = 0\). This equation is satisfied for all \(A > 0\).

Step 12 ：So, the values of \(A\), \(B\), and \(C\) are any positive number, \(-4\) times that number, and \(4\) times that number, respectively. \(\boxed{A > 0, B = -4A, C = 4A}\)