I'm losing my mind with sin...

[jlmatikka]

Hello all! This is my first post and, yes, I read from the "before posting" topic that you shouldn't post list of your homework but I hope you don't mind this time. I really don't proceed on this 'problem' and I have been trying to solve it for hours now. And internet doesn't offer full answer.

So:

3*sin2x = 2*sin4x

There it is! And some background information for you: I took common factor and got x = pi/2 + n*pi as an answer, but those aren't the only solutions. According to my math book, x≈+-0,36 + n*pi are also solutions. How??

-JL

 

[Steven G]

Hello all! This is my first post and, yes, I read from the "before posting" topic that you shouldn't post list of your homework but I hope you don't mind this time. I really don't proceed on this 'problem' and I have been trying to solve it for hours now. And internet doesn't offer full answer.

So:

3*sin2x = 2*sin4x

There it is! And some background information for you: I took common factor and got x = pi/2 + n*pi as an answer, but those aren't the only solutions. According to my math book, x≈+-0,36 + n*pi are also solutions. How??

-JL

You could use the identity that sin2t=2sintcost. So sin4x=sin(2*2x)= 2sin2xcos2x. Then you have 3sin2x=4sin2xcos2x. Set equal to 0 and factor. Show us your new work

 

[jlmatikka]

You could use the identity that sin2t=2sintcost. So sin4x=sin(2*2x)= 2sin2xcos2x. Then you have 3sin2x=4sin2xcos2x. Set equal to 0 and factor. Show us your new work

Thank you for your answer! That's exactly what I did. I took the common factor:
sin2x(3-2cos2x)=0
A: sin2x= 0 or B: 3-2cos2x = 0


A: 2sinxcosx = 0

... sin x = 0 or cos x = 0
x = pi/2 + n*pi (n belongs to Z) or x= pi + n*pi

B: 3-2cos2x= 0 -> no solutions (x belongs to R)

Am I doing something wrong? And what am I missing?

 

[ksdhart]

The problem is the equation you got after factoring is wrong. If you try multiplying the sin(2x) back through, you'll see that:

\(\displaystyle sin\left(2x\right)\cdot \left(3-2\cdot cos\left(2x\right)\right)=0\)
becomes
\(\displaystyle 3\cdot sin\left(2x\right)-2\cdot sin\left(2x\right)\cdot cos\left(2x\right)=0\)

Whereas, you had a 4 in the original equation. So, you should end up with this after factoring:

\(\displaystyle sin\left(2x\right)\cdot \left(3-4cos\left(2x\right)\right)=0\)

 

[jlmatikka]

The problem is the equation you got after factoring is wrong. If you try multiplying the sin(2x) back through, you'll see that:

\(\displaystyle sin\left(2x\right)\cdot \left(3-2\cdot cos\left(2x\right)\right)=0\)
becomes
\(\displaystyle 3\cdot sin\left(2x\right)-2\cdot sin\left(2x\right)\cdot cos\left(2x\right)=0\)

Whereas, you had a 4 in the original equation. So, you should end up with this after factoring:

\(\displaystyle sin\left(2x\right)\cdot \left(3-4cos\left(2x\right)\right)=0\)

That's true! Thank you so much! I became blind with those twos

 

[Steven G]

That's true! Thank you so much! I became blind with those twos

If you think those two's are bad wait till the three's and four's come!