Question

2cos^2(2x) + 3sin4x + 4 sin^2 (2x) = 0

Answer 1

2cos²2x+6sin2xcos2x+4sin²2x=0 /2cos²2x
2tg²2x+3tg2x+1=0
tg2x=a
2a²+3a+1=0
D=9-8=1
a1=(-3-1)/4=-1⇒tg2x=-1⇒2x=-π/4+πn⇒x=-π/8+πn/2
a2=(-3+1)/4=-1/2⇒tg2x=-1/2⇒2x=-arctg1/2+πn⇒x=-1/2arctg1/2+πn/2