Question

2 sin^2 2x - 2 sin 4x + 1= 0

Answer 1

2sin²x-4sin2xcos2x+sin²x+cos²x=0
3sin²x-4sin2xcos2x+cos²x=0 /cos²x≠0
3tg²x-4tgx+1=0
tgx=a
3a²-4a+1=0
D=16-12=4
a1=(4-2)/6=1/3⇒tgx=1/3⇒x=arctg1/3+πn
a2=(4+2)/6=1⇒tgx=1⇒x=π/4+πn