General solution (1 Viewer)

[lacklustre]

Find the general solution of sin3x = sin2x.

 

[Tsylana]

[FONT=Arial,Helvetica,sans-serif]sin(3θ) = 3sin(θ)cos²(θ) - sin³(θ)

[/FONT][FONT=Arial,Helvetica,sans-serif]3sin(θ)cos²(θ) - sin³(θ) = 2Sin[/FONT][FONT=Arial,Helvetica,sans-serif](θ)[/FONT][FONT=Arial,Helvetica,sans-serif]cos(θ)

[/FONT][FONT=Arial,Helvetica,sans-serif]3sin(θ)[1-sin²(θ)] - [/FONT][FONT=Arial,Helvetica,sans-serif]sin³(θ) = [/FONT][FONT=Arial,Helvetica,sans-serif]2Sin[/FONT][FONT=Arial,Helvetica,sans-serif](θ)[/FONT][FONT=Arial,Helvetica,sans-serif]cos(θ)

[/FONT][FONT=Arial,Helvetica,sans-serif]3sin(θ) - [/FONT][FONT=Arial,Helvetica,sans-serif]4sin³(θ) - [/FONT][FONT=Arial,Helvetica,sans-serif]2Sin[/FONT][FONT=Arial,Helvetica,sans-serif](θ)[/FONT][FONT=Arial,Helvetica,sans-serif]cos(θ) = 0

[/FONT][FONT=Arial,Helvetica,sans-serif]4sin³(θ) + [/FONT][FONT=Arial,Helvetica,sans-serif]2Sin[/FONT][FONT=Arial,Helvetica,sans-serif](θ)[/FONT][FONT=Arial,Helvetica,sans-serif]cos(θ) [/FONT][FONT=Arial,Helvetica,sans-serif]- 3sin(θ) = 0

Work from there coz im lazy?
[/FONT]

 

[lacklustre]

[FONT=Arial,Helvetica,sans-serif]sin(3θ) = 3sin(θ)cos<SUP>2</SUP>(θ) - sin<SUP>3</SUP>(θ)

[/FONT]
[FONT=Arial,Helvetica,sans-serif]How did you know this?

[/FONT][FONT=Arial,Helvetica,sans-serif]3sin(θ)cos<SUP>2</SUP>(θ) - sin<SUP>3</SUP>(θ) = 2Sin[/FONT][FONT=Arial,Helvetica,sans-serif](θ)[/FONT][FONT=Arial,Helvetica,sans-serif]cos(θ)

[/FONT][FONT=Arial,Helvetica,sans-serif]3sin(θ)[1-sin<SUP>2</SUP>(θ)] - [/FONT][FONT=Arial,Helvetica,sans-serif]sin<SUP>3</SUP>(θ) = [/FONT][FONT=Arial,Helvetica,sans-serif]2Sin[/FONT][FONT=Arial,Helvetica,sans-serif](θ)[/FONT][FONT=Arial,Helvetica,sans-serif]cos(θ)

[/FONT][FONT=Arial,Helvetica,sans-serif]3sin(θ) - [/FONT][FONT=Arial,Helvetica,sans-serif]4sin<SUP>3</SUP>(θ) - [/FONT][FONT=Arial,Helvetica,sans-serif]2Sin[/FONT][FONT=Arial,Helvetica,sans-serif](θ)[/FONT][FONT=Arial,Helvetica,sans-serif]cos(θ) = 0

[/FONT][FONT=Arial,Helvetica,sans-serif]4sin<SUP>3</SUP>(θ) + [/FONT][FONT=Arial,Helvetica,sans-serif]2Sin[/FONT][FONT=Arial,Helvetica,sans-serif](θ)[/FONT][FONT=Arial,Helvetica,sans-serif]cos(θ) [/FONT][FONT=Arial,Helvetica,sans-serif]- 3sin(θ) = 0

Work from there coz im lazy?
[/FONT]

Hmm lol I'm not sure where to go from there.

 

[Tsylana]

[FONT=Arial,Helvetica,sans-serif]4sin³(θ) + [/FONT][FONT=Arial,Helvetica,sans-serif]2Sin[/FONT][FONT=Arial,Helvetica,sans-serif](θ)[/FONT][FONT=Arial,Helvetica,sans-serif]cos(θ) [/FONT][FONT=Arial,Helvetica,sans-serif]- 3sin(θ) = 0

4[sin^2[/FONT][FONT=Arial,Helvetica,sans-serif](θ)sin[/FONT][FONT=Arial,Helvetica,sans-serif](θ)] + [/FONT][FONT=Arial,Helvetica,sans-serif]2Sin[/FONT][FONT=Arial,Helvetica,sans-serif](θ)[/FONT][FONT=Arial,Helvetica,sans-serif]cos(θ) [/FONT][FONT=Arial,Helvetica,sans-serif]- 3sin(θ) = 0
4[1-cos[/FONT][FONT=Arial,Helvetica,sans-serif]^2(θ)][[/FONT][FONT=Arial,Helvetica,sans-serif]sin[/FONT][FONT=Arial,Helvetica,sans-serif](θ)] + [/FONT][FONT=Arial,Helvetica,sans-serif]2Sin[/FONT][FONT=Arial,Helvetica,sans-serif](θ)[/FONT][FONT=Arial,Helvetica,sans-serif]cos(θ) [/FONT][FONT=Arial,Helvetica,sans-serif]- 3sin(θ) = 0

Factorise out sin theta. Reduce it to a quadratic.

Ummm I learned the triple angle expansions off as a rule just to help me in the the exams... but meh.

Just expand Sin[/FONT][FONT=Arial,Helvetica,sans-serif](2θ+[/FONT][FONT=Arial,Helvetica,sans-serif]θ[/FONT][FONT=Arial,Helvetica,sans-serif])... You should get what i wrote anyway...[/FONT][FONT=Arial,Helvetica,sans-serif][/FONT][FONT=Arial,Helvetica,sans-serif][/FONT]


ok i thought i'd get non-lazy and edit back in the squared signs... but then i pressed the edit button... and got lazy.

 

[lacklustre]

[FONT=Arial,Helvetica,sans-serif]4sin<SUP>3</SUP>(θ) + [/FONT][FONT=Arial,Helvetica,sans-serif]2Sin[/FONT][FONT=Arial,Helvetica,sans-serif](θ)[/FONT][FONT=Arial,Helvetica,sans-serif]cos(θ) [/FONT][FONT=Arial,Helvetica,sans-serif]- 3sin(θ) = 0

4[sin^2[/FONT][FONT=Arial,Helvetica,sans-serif](θ)sin[/FONT][FONT=Arial,Helvetica,sans-serif](θ)] + [/FONT][FONT=Arial,Helvetica,sans-serif]2Sin[/FONT][FONT=Arial,Helvetica,sans-serif](θ)[/FONT][FONT=Arial,Helvetica,sans-serif]cos(θ) [/FONT][FONT=Arial,Helvetica,sans-serif]- 3sin(θ) = 0
4[1-cos[/FONT][FONT=Arial,Helvetica,sans-serif]^2(θ)][[/FONT][FONT=Arial,Helvetica,sans-serif]sin[/FONT][FONT=Arial,Helvetica,sans-serif](θ)] + [/FONT][FONT=Arial,Helvetica,sans-serif]2Sin[/FONT][FONT=Arial,Helvetica,sans-serif](θ)[/FONT][FONT=Arial,Helvetica,sans-serif]cos(θ) [/FONT][FONT=Arial,Helvetica,sans-serif]- 3sin(θ) = 0

Factorise out sin theta. Reduce it to a quadratic.

Ummm I learned the triple angle expansions off as a rule just to help me in the the exams... but meh.

Just expand Sin[/FONT][FONT=Arial,Helvetica,sans-serif](2θ+[/FONT][FONT=Arial,Helvetica,sans-serif]θ[/FONT][FONT=Arial,Helvetica,sans-serif])... You should get what i wrote anyway...[/FONT]


ok i thought i'd get non-lazy and edit back in the squared signs... but then i pressed the edit button... and got lazy.

Wow, thanks a lot. I'll get to work on it then.

 

[Tsylana]

[FONT=Arial,Helvetica,sans-serif]Oh yeah, once you factorise out sin(θ) = 0 [/FONT]
[FONT=Arial,Helvetica,sans-serif]
Solve it as a general solution, then completely erase it from the equation to make it look more clean =].
[/FONT]

 

[Trebla]

sin 3x = sin 2x
Recall that the general solution of sin a = sin b is:
a = kπ + (-1)^k.b (where k is integer valued)
Equivalently, general solution of sin a = c is:
a = kπ + (-1)^k.sin^-1c (where k is integer valued)

So:
3x = kπ + (-1)^k.2x
=> kπ = x[3 - 2.(-1)^k]
=> x = kπ/{3 - 2.(-1)^k}
where k is integer valued

instead of doing all that expansion...