Răspuns:
x∈(8k-1)π/4∪(8k-3)π/4∪(8k-1)π/16∪ (8k+3)π/16
Explicație pas cu pas:
cos5x+sin3x=0
cos5x+cos(π/2-3x)=0
2cos((2x+π/2)/2)cos(4x-π/4)=0
cos (x+π/4)=0
x+π/4 =2kπ+-arccos0=2kπ+-π/2=
x=2kπ+π/2-π/4=2kπ-π/4=(8k-1)π/4
x=2kπ-π/2-π/4=2kπ-3π/4=(8k-3)π/4
(4x-π/4)=0
cu arccos0=π/2 avem
4x=π/4+2kπ ±π/2
x=π/16+kπ/2 ± π/8
x=kπ/2-π/16=(8k-1)π/16
x=kπ/2+3π/16= (8k+3)π/16
x∈(8k-1)π/4∪(8k-3)π/4∪(8k-1)π/16∪ (8k+3)π/16
