Question

$sin4x+sin^{2}2x=0$

Answer 1

sin 4x + sin^2 2x = 0
2sin 2x*cos 2x + sin^2 2x = 0
sin 2x*(2cos 2x + sin 2x) = 0
1) sin 2x = 0; 2x = pi*k; x1 = pi/2*k
2) 2cos 2x + sin 2x = 0
sin 2x = -2cos 2x
tg 2x = -2
2x = -arctg 2 + pi*n;
x2 = -1/2*arctg 2 + pi/2*n

Answer 2

sin 4x + sin² 2x = 0
2sin 2x • cos 2x + sin² 2x = 0
sin 2x • (2cos 2x + sin 2x) = 0
sin 2x = 0; 2x = п • k; x1 = п/2 • k
2cos 2x + sin 2x = 0
sin 2x = - 2cos 2x
tg 2x = - 2
2x = -arctg 2 + п • n;
x2 = -1/2 • arctg 2 + п/2 • n