Question

test for symmetry and then graph the polar equation.
r = 2 + 3\sin\theta
a. is the graph of the polar equation symmetric with respect to the polar axis?
a. yes.
b. the polar equation failed the test for symmetry which means that the graph is not symmetric with respect to the polar axis.
c. the polar equation failed the test for symmetry which means that the graph may or may not be symmetric with respect to the polar axis.
b. is the graph of the polar equation symmetric with respect to the line \theta=\frac{\pi}{2}?
a. the polar equation failed the test for symmetry which means that the graph may or may not be symmetric with respect to the line \theta=\frac{\pi}{2}.
b. yes.
c. the polar equation failed the test for symmetry which means that the graph is not symmetric with respect to the line \theta=\frac{\pi}{2}.

Explanation:

Step1: Test for symmetry about the polar axis ($\theta = 0$)

Replace $\theta$ with $-\theta$.
Original equation: $r = 2+3\sin\theta$.
After replacement: $r = 2 + 3\sin(-\theta)=2-3\sin\theta$.
Since $2 - 3\sin\theta
eq2 + 3\sin\theta$ (in general, unless $\sin\theta = 0$), the equation fails the symmetry - about - polar - axis test. But a failed test does not guarantee lack of symmetry.

Step2: Test for symmetry about the line $\theta=\frac{\pi}{2}$

Replace $\theta$ with $\pi-\theta$.
Original equation: $r = 2+3\sin\theta$.
After replacement: $r=2 + 3\sin(\pi-\theta)$.
Using the identity $\sin(A - B)=\sin A\cos B-\cos A\sin B$ with $A=\pi$ and $B = \theta$, we know that $\sin(\pi-\theta)=\sin\pi\cos\theta-\cos\pi\sin\theta=\sin\theta$.
So $r = 2+3\sin(\pi - \theta)=2 + 3\sin\theta$, which is the same as the original equation.

Answer:

a. C. The polar equation failed the test for symmetry which means that the graph may or may not be symmetric with respect to the polar axis.
b. B. Yes.