Question

sin (2x) - cos (6x)=0

Answer 1

Решение
Sin (2x) - cos (6x)=0
cos(π/2 - 2x) - cos6x = 0
- 2sin[(π/2 - 2x) + 6x]/2 * sin[(π/2 - 2x) - 6x]/2 = 0
sin[(π/2 + 4x)]/2 *  sin[(π/2 - 8x)]/2 = 0
1) sin[(π/2 + 4x)]/2 = 0
 sin(π/4 + 2x) = 0
π/4 + 2x = πk, k ∈ Z
2x = - π/4 + πk, k ∈ Z
x =   - π/8 + π/2, k ∈ Z
2)  sin[(π/2 - 8x)]/2 = 0
sin(π/4 - 4x) = 0
sin(4x - π/4) = 0
4x - π/4 = πn, n ∈ Z
4x = π/4 + πn, n ∈ Z
x = π/16 + πn/4, n ∈ Z