2sinx+3=4 domain is 0<=x<= pi/2
2006-11-12 18:22:37 · 6 answers · asked by John 1 in Science & Mathematics ➔ Mathematics
2sinx+3=4
2sinx=1
sinx=1/2
x = pi/6
2006-11-12 19:02:58 · answer #1 · answered by Anonymous · 1⤊ 0⤋
<=1
2006-11-13 02:25:44 · answer #2 · answered by Anonymous · 0⤊ 1⤋
2sinx+3=4
2sinx=1
sinx=1/2
x=30
2006-11-13 02:34:05 · answer #3 · answered by Chong Sian C 3 · 1⤊ 0⤋
2sin x + 3 =4
=> sin x = 1/2
Note that sin (pi/6) = 1/2
so,
x = n.pi + [(-1)^n].(pi/6)
but since x is between 0 and pi/2 (limits inclusive)
the solution is the principal one, i.e at n=0
x = pi/6
2006-11-13 05:20:05 · answer #4 · answered by yasiru89 6 · 0⤊ 0⤋
2sin x + 3 =4
2sinx=1
sin x = 1/2
Note that
sin (pi/6) = 1/2
cos (pi/6) = sqrt3/2
==> x=sqrt3/2
2006-11-16 02:26:50 · answer #5 · answered by D-Boyz 2 · 0⤊ 0⤋
2sinx+3=4
2sinx =1
sinx = 1/2
x = 30+360*k (k=....,-3,-2,-1,0,1,2,3,.....)
2006-11-13 03:15:10 · answer #6 · answered by uknowhu 1 · 0⤊ 0⤋