Question

2cos2(6x)+8sin(3x)cos(3x)-4=0​

Answer 1

$2cos^2(6x)+8sin(3x)cos(3x)-4=0\\2(1-sin^2(6x))+4sin(6x)-4=0\\sin(6x)=t; -1\leq t\leq 1\\2(1-t^2)+4t-4=0 |:(-2)\\-1+t^2-2t+2=0\\t^2-2t+1=0\\(t-1)^2=0\\t-1=0\\t=1\\sin(6x)=1\\6x=\frac{\pi}{2} +2\pi k, k \in Z\\x=\frac{\pi}{12} +\frac{\pi k}{3}, k \in Z$