solve(in radians) for values of x between 0 and 2pi rad 1. sin(x)-1=0 2. 4cos^2x-1=0?
回答 (3)
2016-11-28 5:41 pm
1) x = π/2
2) cos x = ±1/2
x = π/3, 2π/3, 4π/3, 5π/3
2) cos x = ±1/2
x = π/3, 2π/3, 4π/3, 5π/3
2016-11-28 8:56 pm
sinx=1+.2=1.2(impossible)
(2cosx-1)(2cosx+1)=0
2cosx-1=0,,2cosx=1,,,cosx=1/2 then x=pi/3,5pi/3
2cosx+1=0 then cosx=-1/2,x=2pi/3,4pi/3
(2cosx-1)(2cosx+1)=0
2cosx-1=0,,2cosx=1,,,cosx=1/2 then x=pi/3,5pi/3
2cosx+1=0 then cosx=-1/2,x=2pi/3,4pi/3
2016-11-28 7:33 pm
1.
sin x = 1
x = π / 2
2.
cos²x = 1/4
cos x = ± 1/2
x = π/3 , 2π/3 , 7π/6 , 11π/6
sin x = 1
x = π / 2
2.
cos²x = 1/4
cos x = ± 1/2
x = π/3 , 2π/3 , 7π/6 , 11π/6
收錄日期: 2021-04-23 23:57:36
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20161128093415AA5ZDGZ