Melody

Welcome as a member Tiggsy.

I should have said this immediately but I did not realize straight away that you had joined!

I know you have been here forever but I finally talked you into becoming a member.  I am very pleased! 

Now I will be able to look back and find your fabulous answers (from this point forward)

Finally ... I have exhausted this question.  There was a lot for me to work through here. 

I do not remember seeing those formulas before that you used Tiggsy. Hopefully now they, and their proofs, are a little embedded into my brain.

Anyway, i finally fully understand both solutions so I am happy.      

The 3rd answer is the right one.

You can see that 2 should occur about twice as often as each of the other 2.

This is a NET!

You are both taking the long route   !

a) What does t = in the year 2000?    Sub that in.

b) exponential  

c) it is 2019 now.  What is t?  Sub that in.

First

You take the first derivative and put it equal to 0

Have you done that?

What did you get?

Nicely answered HiylinLink    

Thanks Chris, I didn't realize there were 2 sets of solutions.     

Hello there OldTimer     

It is nice to see you again.

\(Sin3xCos3x - Cos^2 2x + 1/2 = 0\\ 2Sin3xCos3x - 2Cos^2 2x + 1 = 0\\ sin(6x) - 2Cos^2 2x + 1 = 0\\ \qquad cos4x=cos^22x-sin^22x\\ \qquad cos4x=cos^22x-(1-cos^22x)\\ \qquad cos4x=2cos^22x-1\\ \qquad -cos4x=-2cos^22x+1\\ so\\ sin(6x) - 2Cos^2 2x + 1 = 0\\ sin(6x) - cos(4x)= 0\\\)

\(sin(6x) =cos(4x)\)

NOW

\(\qquad cos(\theta)=sin(\frac{\pi}{2}-\theta)=sin(\pi-(\frac{\pi}{2}-\theta))=sin(\frac{\pi}{2}+\theta)\\ \qquad \therefore\;\;\;\;cos(4x)=sin(\frac{\pi}{2}+4x)\)

\(sin(6x)=cos(4x)\\ sin(6x)=sin(\frac{\pi}{2}+4x)\\ 6x=\frac{\pi}{2}+4x+2\pi n \qquad n\in Z\\ 2x=\frac{\pi}{2}+2\pi n \qquad n\in Z\\ x=\frac{\pi}{4}+\pi n \qquad n\in Z\\\)

\({(\sqrt{2})^{8}(\sqrt[4]{4})^{4}}\\ =2^4*4^1\\ =4^2*4^1\\ =4^3\\ \)