Найдите производную функций f(x)=sinx-cosx f(x)=4x-sinx f(x)=tgx-4ctgx f(x)=6cosx-1,2x
264
461
Ответы на вопрос:
1) cos(x/2) = 0 x/2 = π/2 + πk, k∈z x = π + 2πk, k∈z 2) 4 + 3cos2x = 1 3cos2x = - 3 cos2x = - 1 2x = π + 2πn, n∈z x = π/2 + πn, n∈z 3) cos(6+3x) = - ( )6 + 3x = (+ -)arccos(- √2/2) + 2πk, k∈z 6 + 3x = (+ π - arccos√2/2) + 2πk, k∈z 6 + 3x = (+ -)*(π - π/4) + 2πk, k∈z 6 + 3x = (+ π/4) + 2πk, k∈z 3x = (+ -)*(3π/4) - 6 + 2πk, k∈z x = (+ -)*(π/4) - 2 + 2πk/3, k∈z
Популярно: Алгебра
-
vdimitrievap0donn10.01.2022 03:20
-
Ааааааааввв04.03.2022 22:30
-
12344312324.10.2021 19:21
-
юля1510228.09.2020 22:08
-
egordyachenko15.07.2020 02:16
-
Дзера111131.10.2022 01:31
-
GrootShkolnik02.06.2022 06:40
-
alyonasajko24.03.2023 13:29
-
KaPitanKartoIIIka09.04.2021 21:47
-
uhsdjhsilhiilch24.10.2020 05:33