One trig problem?
回答 (3)
2016-06-15 1:56 pm
5 cot²(x) = 5
cot²(x) = 1
cot(x) = 1 or cot(x) = -1
x = π/4, π + (π/4) or x = π - (π/4), 2π - (π/4)
x = π/4, 3π/4, 5π/4, 7π/4
cot²(x) = 1
cot(x) = 1 or cot(x) = -1
x = π/4, π + (π/4) or x = π - (π/4), 2π - (π/4)
x = π/4, 3π/4, 5π/4, 7π/4
2016-06-15 1:53 pm
5cot²(x) = 5
cot²(x) = 1
cot(x) = ±1
cos(x)/sin(x) = ±1
cos(x) = ±sin(x)
x = π/4, 3π/4, 5π/4, 7π/4
cot²(x) = 1
cot(x) = ±1
cos(x)/sin(x) = ±1
cos(x) = ±sin(x)
x = π/4, 3π/4, 5π/4, 7π/4
2016-06-15 4:05 pm
tan²x = 1
tan x = ± 1
x = π/4 , 3π/4 , 5π/4 , 7π/4
tan x = ± 1
x = π/4 , 3π/4 , 5π/4 , 7π/4
收錄日期: 2021-04-18 15:08:28
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20160615055114AAxUpxL