Question

2cos^2(2x)+3sin(4x)+4sin^2(2x)=0

Answer 1

2*(1+сos4x)/2 *3sin4x+4*(1-cos4x)/2=0
1+cos4x+3sin4x+2-2cos4x=0
1-cos4x+3sin4x=0
2sin²2x+6sin2xcos2x=0/2cos²2x≠0
tg²2x+3tg2x=0
tg2x(tg2x+3)=0
tg2x=0⇒2x=πn⇒x=πn/2
tg2x=-3⇒2x=-arctg3+πn⇒x=-1/2arctg3+πn/2