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Solve:  Sin3xCos3x - Cos^2 2x + 1/2 = 0 

 

Ans:  n x 36 deg + 9 deg , 45 deg - n x 180

 Apr 7, 2019
 #1
avatar+107115 
+4

Hello there OldTimer     laugh

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

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


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

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 Apr 7, 2019
 #2
avatar+106539 
+1

Nice solution,  Melody  !!!!!

 

To get the other solution, we have

 

sin(6x)  = cos(4x)

 

cos( pi/2 - 6x) = cos(4x)

 

pi/2 - 6x + 2pi * n   =  4x

 

pi/2 + 2pi * n =  10x

 

x = pi/20 + pi /5 *n

 

x = 9° + (36°) * n

 

 

cool cool cool

 Apr 7, 2019
edited by CPhill  Apr 7, 2019
 #3
avatar+107115 
+2

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

Melody  Apr 7, 2019
 #4
avatar+160 
+2

Thanks a lot Melody & CPHill - you've got talent!...... & you have restored my sanity!

 Apr 8, 2019
edited by OldTimer  Apr 8, 2019

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