Thread: Trig Equation

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?

Hello, kinhew93!

Your substitution is incorrect: .

I would solve it like this . . .

n . . . . . . . . .

. . . . .

Factor: .


. .

. .

Originally Posted by Soroban

Hello, kinhew93!

Your substitution is incorrect:

In this case it is because you can times through by sinA...

Thanks though you were a great help! I knew I was make it far to complicated!