sin4x+sin2x=0 find the general solution
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sin 4x - sin 2x = 0
2 sin 2x cos 2x - sin 2x = 0
(2 cos 2x - 1) sin 2x = 0
Zero product property.
Case 1: 2 cos 2x - 1 = 0
cos 2x = 1/2
2x = 2kπ ± π/3 ; k ∈ Z
x = kπ ± π/6 ; k ∈ Z
Case 2: sin 2x = 0
2x = kπ ; k ∈ Z
x = kπ/2 ; k ∈ Z
Thus the full solution set is:
{kπ/2 , kπ±π/6 | k ∈ Z}
Within the 0<x<2π range, this means:
{π/6,π/2,5π/6,π,7π/6,3π/2,11π/6}
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