1. ## 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?

2. ## Re: Trig Equation

Hello, kinhew93!

Your substitution is incorrect: .$\displaystyle \sin A \:=\:\sqrt{1-\cos^2\!A}$

$\displaystyle 3\sin2A \:=\: \sin A$

I would solve it like this . . .

n . . . . . . . . . $\displaystyle 6\sin A\cos A \:=\:\sin A$

. . . . .$\displaystyle 6\sin A\cos A - \sin A \:=\:0$

Factor: .$\displaystyle \sin A(6\cos A - 1) \:=\:0$

. . $\displaystyle \sin A \:=\:0 \quad\Rightarrow\quad \boxed{A \:=\:\pi n}$

. . $\displaystyle 6\cos A - 1 \:=\:0 \quad\Rightarrow\quad \cos A \:=\:\tfrac{1}{6} \quad\Rightarrow\quad \boxed{A \:=\:\cos^{-1}\left(\tfrac{1}{6}\right) + 2\pi n}$

3. ## Re: Trig Equation

Originally Posted by Soroban
Hello, kinhew93!