Exercise 71 Page 918

Given $a>0,$ we want to determine the conditions for $a$ and $b$ so that the function $a\sin \theta =b$ to has exactly two solutions in the interval $0\le \theta <2\pi .$ To do so, let's first isolate $\sin \theta$ using the Division Property of Equality. Now we can see that we are actually looking for values of a standard, non-transformed sine function. The range of a sine function is between $\text{-}1$ and $1.$

Because we want to only consider the values of $y=\sin \theta$ that are equal to two different values of $\theta ,$ we need to look at every point in this interval except the maximum and the minimum, $1$ and $\text{-}1.$ For both $\sin \theta =1$ and $\sin \theta =\text{-}1,$ there is only one possible value of $\theta .$

Since the maximum and minimum values would not have two solutions in the given interval, we know that the values of $\sin \theta$ must be between, and not including, $\text{-}1$ and $1.$ Now, we can substitute $\frac{b}{a}$ for $\sin \theta$ in this inequality. Finally, using the Multiplication Property of Inequality, we can multiply each side by $a.$ We were given that $a$ is positive so we do not need to worry about flipping the inequality sign. Therefore, if $b$ is less than $a$ but greater than $\text{-}a,$ the given function has exactly two solutions. This corresponds to option D.