# 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.