Question

1)решите неравенство: cost ≤ 1/3
2)найдите корни уравнения сos(x+pi/12)=0 на промежутке [0;3pi/2]

Answer 1

1
cosx≤1/3
x∈[arccos1/3+2πk;2π-arccos1/3+2πk,k∈z]
2
cos(x+π/12)=0
x+π/12=π/2+πk
x=-π/12+π/2+πk
x=5π/12+πk,k∈z
0≤5π/12+πk≤3π/2
0≤5+12k≤18
-5≤12k≤13
-513≤k≤13/12
k=0⇒x=5π/12
k=1⇒x=5π/12+π=17π/12