Hello, kinhew93!
Your substitution is incorrect: .
I would solve it like this . . .
n . . . . . . . . .
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Factor: .
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3sin(2A) = sin(A)
There are a number problems like this in my textbook and I am able to solve them, but I always end up with a polynomial in some trig function. I wanted to know if there were any less time consuming approaches to this question?
Here's what I did for the above example:
3sin(2A) = sin(A)
6sin(A)cos(A) = sin(A)
6(1-cos^2(A))(cos(A)) = 1 - cos^2(A)
6cos^3(A) - cos^2(A) - 6cos(A) + 1 = 0
and then solved to find cos(A).
I have a feeling that I am making the question more time-consuming than neccessary - is there a simpler approach?