We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website. cookie policy and privacy policy.
 
+0  
 
0
39
4
avatar+115 

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+100006 
+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

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


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

.
 Apr 7, 2019
 #2
avatar+99237 
+3

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+100006 
+2

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

Melody  Apr 7, 2019
 #4
avatar+115 
+2

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

 Apr 8, 2019
edited by OldTimer  Apr 8, 2019

29 Online Users

avatar
avatar