1)y=sin2nx ; 2)y=|cosx|; 3)y=1+sin2x; 4)y=sin2x+3cos3x; 5)y=tg3x+5ctg2x. - id1650913320230206 от yxhcycgc 06.02.2023 13:44

Question

Алгебра: 1)y=sin2nx ; 2)y=|cosx|; 3)y=1+sin2x; 4)y=sin2x+3cos3x; 5)y=tg3x+5ctg2x.​

yxhcycgc · Accepted Answer

Решение 1)  2cosx-1 0 cosx 1/2 arccos(1/2) + 2πn x 2π - arccos(1/2) + 2πn, n ∈ Z π/3 + 2πn x 2π - π/3 + 2πn, n ∈ Z π/3 + 2πn x  5π/3 + 2πn, n ∈ Z 2)  sin2x - √2/2   0  sin2x   √2/2  - π - arcsin(√2/2) + 2πk 2x arcsin(√2/2) + 2πk, k ∈ Z - π - π/4 + 2πk 2x π/4 + 2πk, k ∈ Z  - 5π/4 + 2πk 2x π/4 + 2πk, k ∈ Z  - 5π/8 + πk x π/8 + πk, k ∈ Z 3)  tgx 1 - π/2 + πn x arctg(1) + πn, n ∈ Z - π/2 + πn x π/4 + πn, n ∈ Z...

1)y=sin2nx ;2)y=|cosx|;3)y=1+sin2x;4)y=sin2x+3cos3x;5)y=tg3x+5ctg2x.​

MOGZ ответил

1)y=sin2nx ;
2)y=|cosx|;
3)y=1+sin2x;
4)y=sin2x+3cos3x;
5)y=tg3x+5ctg2x.