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

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

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

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

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

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