5sin@ -2cos^2@-1 =0, 0度 <_ @<_ 360度 (2個都係大過等於)
點計?
中4一問maths
2009-09-10 5:56 am
回答 (1)
2009-09-10 6:07 am
✔ 最佳答案
5sin@ -2cos^2@-1 =0
5sin@ - 2(1 - sin^2@) - 1 = 0
2sin^2 @ + 5sin@ - 3 = 0
(2sin@ - 1)(sin@ + 3) = 0
2sin@ - 1 = 0 or sin@ + 3 = 0
sin@ = 1/2 or sin@ = - 3 (Since -1<= sinx <=1 , rejected.)
From sin@ = 1/2
@ = 30度 , 180-30 = 150度 for 0度 <_ @<_ 360度
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原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20090909000051KK01743