General solution of trig functions

[BOAS]

Hello,

Homework Statement

find the general solution to cos3θ = sin2θ

Homework Equations

The Attempt at a Solution

I know that sinθ = cos(π/2 - θ) but I am unsure of how to apply this when I have sin2θ.

Do I say that sin2θ = cos2(π/2 - θ)?

I think not because when I do this my answer does not agree with what my book says.

Thanks!

 

[BOAS]

I thought i'd resolved my problem, but it still does not agree with the book.


sinθ = cos(π/2 - θ)

sin2θ = cos(π/2 - 2θ)

cos3θ = cos(π/2 - 2θ)

3θ = nπ +/- π/2 - 2θ

θ = (nπ +/- π/2 - 2θ)/3

 

[BOAS]

I feel like a crazy person talking to myself, but just to save anyone the trouble of replying, I have figured it out.

3θ = nπ +/- π/2 - 2θ

either 3θ = nπ - π/2 - 2θ

or

3θ = nπ + π/2 - 2θ

It works out that θ = π/10(4n+1)

or

θ = π/2(4n-1)

Which is precisely what I wanted :)

Perhaps a mod can delete this thread.

 

[Ray Vickson]

I feel like a crazy person talking to myself, but just to save anyone the trouble of replying, I have figured it out.

3θ = nπ +/- π/2 - 2θ

either 3θ = nπ - π/2 - 2θ

or

3θ = nπ + π/2 - 2θ

It works out that θ = π/10(4n+1)

or

θ = π/2(4n-1)

Which is precisely what I wanted :)

Perhaps a mod can delete this thread.

I hope a moderator does not delete this thread, because you have certainly not found the general solution.

Applying $\sin(\phi) = \cos(\pi/2 \: - \: \phi)$ to $\phi = 2 \theta$, we have
\sin(2 \theta) = \cos\left( \frac{\pi}{2} - 2 \theta \right)
for what it's worth (which is not much in this problem).

 

[BOAS]

I hope a moderator does not delete this thread, because you have certainly not found the general solution.

Applying $\sin(\phi) = \cos(\pi/2 \: - \: \phi)$ to $\phi = 2 \theta$, we have
\sin(2 \theta) = \cos\left( \frac{\pi}{2} - 2 \theta \right)
for what it's worth (which is not much in this problem).

My answer agrees with the textbook... Is what I've done not what you'd call the general solution to the original equation, or are you saying I have made a mistake?

Thanks.

 

[Ray Vickson]

My answer agrees with the textbook... Is what I've done not what you'd call the general solution to the original equation, or are you saying I have made a mistake?

Thanks.

No, I'm saying I made a dumb mistake, and your solution is OK. Sorry.