The range of a standard, non-transformed sine function y=sin θ is between - 1 and 1 in the interval 0≤θ<2π.
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Given a>0, we want to determine the conditions for a and b so that the function asinθ=b to has exactly two solutions in the interval 0≤θ<2π. To do so, let's first isolate sinθ using the Division Property of Equality.
asinθ=b ⇒ sinθ=b/a
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 - 1 and 1.
Because we want to only consider the values of y=sinθ that are equal to two different values of θ, we need to look at every point in this interval except the maximum and the minimum, 1 and -1. For both sinθ=1 and sinθ=-1, there is only one possible value of θ.
Since the maximum and minimum values would not have two solutions in the given interval, we know that the values of sinθ must be between, and not including, -1 and 1.
- 1< sinθ <1
Now, we can substitute ba for sinθ in this inequality.
- 1< sinθ <1 ⇔ - 1< b/a <1
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.
