Hence,solve 2sin^2(90̊ - x)-1/sin(90̊+x)+sinx=20sinxtanx for 0° <θ< 360.
(give your answers correct to 1 decimal place)
已知2sin^2(90̊ - x)-1/sin(90̊+x)+sinx=cosx-sinx
f4 maths
2010-05-23 5:53 am
回答 (2)
2010-05-23 6:38 am
✔ 最佳答案
cos x - sin x = 20 sin x tan xcos x - sin x = 20 sin^2 x/cos x
cos^2 x - sin x cos x = 20 sin^2 x
20 sin^2 x + sin x cos x - cos^2 x = 0
(4 sin x + cos x)(5 sin x - cos x) = 0
so 4 sin x + cos x = 0 ........(1) and
5 sin x - cos x = 0..............(2)
From (1),
4 sin x = - cos x
tan x = -1/4
x = arctan (-1/4) = 346 degree or 166 degree.
From (2),
5 sin x = cos x
tan x = 1/5
x = arctan (1/5) = 11.3 degree or 191.3 degree.
2010-05-22 22:41:51 補充：
Note: It should be [2 sin^2 ( 90 - x) - 1]/[ sin ( 90 + x) + sin x ] = cos x - sin x
2010-05-23 6:02 am
2sin^2(90̊ - x)-[1/sin(90̊+x)]+sinx=cosx-sinx ??
收錄日期: 2021-04-22 00:46:54
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20100522000051KK01626