Mastering Double Angle Identities: Solving -2sin3θ+1=0

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The discussion revolves around solving the equation 8 sin3 θ – 6 sin θ + 1 = 0, which can be transformed into -2sin3θ + 1 = 0. Participants emphasize the importance of understanding the difference between identities and conditional equations, clarifying that this problem is about solving an equation rather than proving an identity. Suggestions include using the addition formula for sine, specifically sin(3θ) = sin(2θ + θ), to aid in the solution process. The conversation highlights the need for careful expansion and manipulation of the equation to find the correct approach. Ultimately, the focus is on mastering the application of double angle identities and addition formulas in solving trigonometric equations.
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Homework Statement


8 sin3 θ – 6 sin θ + 1 = 0

The answer includes changing this to
-2sin3θ+1=0

Homework Equations


The double angle identities
Sin2θ=sinθcosθ+cosθsinθ

The Attempt at a Solution



I do not know how to get started with this question
 
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lioric said:

Homework Statement


8 sin3 θ – 6 sin θ + 1 = 0

The answer includes changing this to
-2sin3θ+1=0

Homework Equations


The double angle identities
Sin2θ=sinθcosθ+cosθsinθ

The Attempt at a Solution



I do not know how to get started with this question

You could try looking at ##\sin(3\theta) = \sin(2\theta + \theta)##.
 
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PeroK said:
You could try looking at ##\sin(3\theta) = \sin(2\theta + \theta)##.
I am familiar with that
Which is the double angle identity
But I fail to see the the relation
I believe that I am missing the transition identity which brings them together
Could you point it out for me please
 
lioric said:
I am familiar with that
Which is the double angle identity
But I fail to see the the relation
I believe that I am missing the transition identity which brings them together
Could you point it out for me please

Expand the expression in post #2. And then have another think.
 
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PeroK said:
Expand the expression in post #2. And then have another think.
This what I came to when I expanded
What do I do from there
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lioric said:
This what I came to when I expanded
What do I do from there
View attachment 221431

So, what do you get for ##2 \sin 3\theta ?##
 
Ray Vickson said:
So, what do you get for ##2 \sin 3\theta ?##
Oh
You mean twice that huh ok let me see
 
lioric said:
Oh
You mean twice that huh ok let me see
Thank you got it now
Thank you very much
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lioric said:

Homework Statement


8 sin3 θ – 6 sin θ + 1 = 0

The answer includes changing this to
-2sin3θ+1=0

Homework Equations


The double angle identities
Sin2θ=sinθcosθ+cosθsinθ

The Attempt at a Solution



I do not know how to get started with this question
It can't be an identity, because when I substitute 90 degrees for theta it comes out to - 8 - 6 + 1 = 0. What is it you are supposed to do? Solve for theta?
 
  • #10
PeroK said:
You could try looking at ##\sin(3\theta) = \sin(2\theta + \theta)##.

lioric said:
I am familiar with that
Which is the double angle identity
No, PeroK is setting up to use an addition formula, not a double angle identity.

You seem to be confused about the concepts of proving an identity versus solving a conditional equation. This problem, which you mistakenly titled "I need help with this identity," actually entails solving an equation.

The difference between these two kinds of equations is that an identity is true for all values of the variable; for example, ##\sin^2(\theta) + \cos^2(\theta) = 1##. A conditional equation is one that is true for a limited number of values of the variable; for example ##x^2 - 2x + 1 = 0##.
 
  • #11
lioric said:
8 sin3 θ – 6 sin θ + 1 = 0

The answer includes changing this to
-2sin3θ+1=0
Only tangentially related (pun intended), but this problem has interesting historical importance: https://en.wikipedia.org/wiki/Angle_trisection
 

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