Homework Help: General solution of trig functions

1. Jan 2, 2014

BOAS

Hello,

1. The problem statement, all variables and given/known data

find the general solution to cos3θ = sin2θ

2. Relevant equations

3. 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!

2. Jan 2, 2014

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

3. Jan 2, 2014

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.

4. Jan 2, 2014

Ray Vickson

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).

5. Jan 2, 2014

BOAS

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.

6. Jan 2, 2014

Ray Vickson

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