How can I find angles that satisfy 3sin(2x) = sin(x)?

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The equation 3sin(2x) = sin(x) led to initial solutions of 80.4 degrees and 279.6 degrees, but the user sought to find additional solutions including 0, 180, and 360 degrees. It was noted that dividing both sides by sin(x) could eliminate potential solutions, prompting a correction in the approach. The correct method involves factoring out sin(x) after rearranging the equation. Ultimately, the user learned to derive the additional angles by setting the factored equation sin(6cos(x) - 1) = 0. This discussion highlights the importance of careful manipulation of trigonometric equations to uncover all possible solutions.
Peter G.
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Hi,

3sin(2x) = sin (x)

I managed to find two answers: 80.4 Degrees and 279.6 degrees but I don't know hot to get 0, 180 and 360 as answers, can anyone help me?

This is how I found the two other angles:

6 sin (x) * cos (x) = sin (x)
cos (x) = sin (x) / 6 sin (x)
cos (x) = 1/6

Thanks in advance,
Peter G.
 
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What can you say about what sin (0), sin (180) and sin (360) have in common?
 
All of them are equal to 0.
 
So, could any of those angles satisfy the original equation?
 
Peter G. said:
Hi,

3sin(2x) = sin (x)

I managed to find two answers: 80.4 Degrees and 279.6 degrees but I don't know hot to get 0, 180 and 360 as answers, can anyone help me?

This is how I found the two other angles:

6 sin (x) * cos (x) = sin (x)
cos (x) = sin (x) / 6 sin (x)

You shouldn't have divided both sides by sin(x). You lose potential solutions that way. Instead, subtract sin(x) from both sides and factor out the greatest common factor (sin(x)).
 
Hi,

Thanks guys. With my teacher and your input, I understood what I did wrong:

6 sin (x) * cos (x) = sin (x)
6 sin (x) * cos (x) - sin (x) = 0

sin (6cos(x) - 1) = 0
 
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Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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