2007-06-29 14:50:01 · 5 answers · asked by joseph j 1 in Science & Mathematics ➔ Mathematics
3sinA=2cos^2A
Evaluating:
3 cosA = 2 * (2 cosA) * (-sinA)
3 cosA = -4 cosA sinA
3 = -4 sinA
sinA = -3/4
By arcsin function:
A = -48.59 degrees (or 311.49 degrees)
In radians (1 degree = 0.0174532925 radian):
A = -0.848 radians (or 5.44 radians)
2007-06-29 15:05:49 · answer #1 · answered by pisayweb 3 · 0⤊ 0⤋
3sinA=2cos^2A
3sinA = 2(1-sin^2A)
3sinA = 2-2sin^2A
2sin^2A + 3sinA -2 = 0
(2sinA - 1)(sinA + 2) =0
2sinA- 1=0 --> sinA = 1/2 --> A = pi/6 , 5pi/6 [0
2007-06-29 22:11:06 · answer #2 · answered by ironduke8159 7 · 0⤊ 0⤋
Use cos^2A=1-sin^2A
3sinA=2-2sin^2A
2sin^2A+3sinA-2=0
(2sinA-1)(sinA+2)=0
sinA=1/2 or sinA=-2
If sinA=1/2 A=Pi/6 + 2kPi and A=5Pi/6 + 2kPi, k is an integer
sinA=-2 has no solution.
2007-06-29 22:06:45 · answer #3 · answered by Anonymous · 0⤊ 0⤋
3sinA= 2COS^a
3sinA =2(1-sin^2A)
3sinA=2-2sin^2A
2sin^2A +3sinA -2=0
(2sinA-1)(sinA+2)=0
sinA =-2,1/2
2007-06-29 22:33:37 · answer #4 · answered by gerrythelips 1 · 0⤊ 0⤋
3 sin A = 2.(1 - sin²A)
3.sin A = 2 - 2.sin²A
2.sin²A + 3.sin A - 2 = 0
sin A = [- 3 ± â25] / 4
Sin A = 1/2 , sin A = - 8 / 4 = - 2 (not permissable)
A = Ï/6 , A = 5Ï/6
2007-06-30 14:16:46 · answer #5 · answered by Como 7 · 0⤊ 0⤋