f.4 trigonometric
2007-03-05 3:07 am
if the quadratic equation 3x^2 - (2sin A)x + cos 2A =0 has equal root,find the possible values of A,correct the answers to the nearest degree.
回答 (2)
2007-03-05 3:15 am
✔ 最佳答案
discriminant = [-(sin A)]^2 - 4(3)(cos 2A) = 0 [because of equal roots](sin A)^2 - 12[1 - 2(sin A)^2] = 0
25(sin A)^2 = 12
A = 44 degrees or A = 136 degrees, correct to the nearest degree
[where A lies between 0 degree and 360 degrees]
2007-03-05 3:20 am
3x^2 - (2sin A)x + cos 2A =0
discriminant=[-(2sinA)]^2 - 4(3)(cos2A)=0
4sin^2A - 12cos2A=0
4sin^2 A - 12(1-sin^2 A)=0
4sin^2 A - 12 + 12sin^2A=0
16sin^2 A=12
sin^2 A=3/4
sinA=root3/2 or sinA= -root3/2
A=60 or -60
I hope I didn't make any mistakes,,,, as I am not good at doing calculation in computer...^^
discriminant=[-(2sinA)]^2 - 4(3)(cos2A)=0
4sin^2A - 12cos2A=0
4sin^2 A - 12(1-sin^2 A)=0
4sin^2 A - 12 + 12sin^2A=0
16sin^2 A=12
sin^2 A=3/4
sinA=root3/2 or sinA= -root3/2
A=60 or -60
I hope I didn't make any mistakes,,,, as I am not good at doing calculation in computer...^^
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