Find exact value 2sin(θ)=2cos(2θ)?
回答 (3)
2016-07-16 5:41 am
2sin(x) = 2cos(2x)
2sin(x) = 2* 1-2sin^2(x)
2sin(x) = 2 -4sin^2(x)
4sin^2(x) + 2sin(x) -2 =0
(2 sin(x) + 2)(2 sin(x) - 1) = 0
sin(x) = -1 sin(x) = 1/2
x = 3pi/2 + 2pi, pi/6 + 2pi, 5pi/6 +2pi
2sin(x) = 2* 1-2sin^2(x)
2sin(x) = 2 -4sin^2(x)
4sin^2(x) + 2sin(x) -2 =0
(2 sin(x) + 2)(2 sin(x) - 1) = 0
sin(x) = -1 sin(x) = 1/2
x = 3pi/2 + 2pi, pi/6 + 2pi, 5pi/6 +2pi
2016-07-15 10:37 pm
cos 2Ө - sin Ө = 0
1 - 2 sin² Ө - sin Ө = 0
2 sin² Ө + sin Ө - 1 = 0
[ 2 sin Ө - 1 ] [ sin Ө + 1 ] = 0
sin Ө = 1/2 , sin Ө = - 1
Ө = 30⁰ , 150⁰ , 270⁰
1 - 2 sin² Ө - sin Ө = 0
2 sin² Ө + sin Ө - 1 = 0
[ 2 sin Ө - 1 ] [ sin Ө + 1 ] = 0
sin Ө = 1/2 , sin Ө = - 1
Ө = 30⁰ , 150⁰ , 270⁰
2016-07-15 7:58 pm
Ask ur teacher
收錄日期: 2021-04-21 19:18:35
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20160715115802AAch4GX