Solve 2cos2x =sins-1?
回答 (1)
2017-09-26 2:56 pm
The question should be: 2 cos(2x) = sin(x) - 1 instead.
Identity used: cos(2x) = 1 - 2 sin²(x)
2 cos(2x) = sin(x) - 1
2 [1 - 2 sin²(x)] = sin(x) - 1
2 - 4 sin²(x) = sin(x) - 1
4 sin²(x) + sin(x) - 3 = 0
[4 sin(x) - 3] [sin(x) + 1] = 0
sin(x) = 3/4 or sin(x) = -1
x = 48.59°, (180-48.59)° or x = 270° (for 0° ≤ x ≤ 360°)
Hence, x = 48.59°, 131.41°, 270°
Identity used: cos(2x) = 1 - 2 sin²(x)
2 cos(2x) = sin(x) - 1
2 [1 - 2 sin²(x)] = sin(x) - 1
2 - 4 sin²(x) = sin(x) - 1
4 sin²(x) + sin(x) - 3 = 0
[4 sin(x) - 3] [sin(x) + 1] = 0
sin(x) = 3/4 or sin(x) = -1
x = 48.59°, (180-48.59)° or x = 270° (for 0° ≤ x ≤ 360°)
Hence, x = 48.59°, 131.41°, 270°
收錄日期: 2021-04-18 17:55:13
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20170926064452AA4qnhj