solve the following equations for 0<x<2兀 (give your answer correct to 3
significant figures.)
3sin((x/3)+1)-2=0
書後答案:4.24rad
但係我計極都係:2.1368rad
prove that (cos@-sec@)^2=sec^2@-sin^2@-1
hence, solve the equation sec^2@=sin^2@+1 for 0<@<2兀
(leave your answers in term of 兀 if necessary)
trigonometric function
2010-05-10 6:36 am
回答 (2)
2010-05-10 7:19 am
✔ 最佳答案
sec^2@=sin^2@+1 for (0 < @ < 2 TT)3sin((x/3)+1)-2=0
sin( x/3 + 1 ) = 2/3
x/3 + 1 = arcsin (2/3)
x/3 + 1 = n TT + [(-1)^n] 0.72973
x = 3 { n TT - 1 + [(-1)^n] 0.72973 }
When n = 0 ,
x = 3 (- 1 + 0.72973) < 0 rejected
When n = 1 ,
x = 3(TT - 1 - 0.72973) = 4.23559 = 4.24(2dec.)
When n = 2 ,
x = 3 (2 TT - 1 + 0.72973) = 18.03875 rejected
prove that (cos@-sec@)^2=sec^2@-sin^2@-1
(cos@-sec@)^2
= cos^2@ - 2 + sec^2@
= sec^2@ + cos^2@ - 1 - 1
= sec^2@ - sin^2@ - 1
= sec^2@-sin^2@-1
solve the equation sec^2@=sin^2@+1 for 0<@<2兀
sec^2@-sin^2@-1= 0
(cos@-sec@)^2= 0
cos@ = sec@
cos^2@ = 1
cos@ = 1 or - 1
@ = 0 (rejected) or TT
@ = TT for (0 < @ < 2 TT)
2010-05-11 1:08 am
收錄日期: 2021-04-21 22:12:01
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https://hk.answers.yahoo.com/question/index?qid=20100509000051KK01677