trigonometry
回答 (2)
2008-01-20 8:19 pm
✔ 最佳答案
sin a = x + 1/xx^2 - (sin a) x + 1 = 0.
discriminant = (-sin a)^2-4 = (sin^2) a - 4
Since -1<=sin a <=1, (sin^2) a <=1,
hence discriminant <= -3 < 0.
Therefore x has no real root.
2008-01-20 6:22 am
Let f(x)=x+1/x
By equating f'(x)=1-1/x^2=0, x=1
Evaluating f"(x)=2/x^3=2>0,
We find that f(1)=2 is a minimum.
Since -1<=sin(x)<=1 for all real x, x cannot be real.
By equating f'(x)=1-1/x^2=0, x=1
Evaluating f"(x)=2/x^3=2>0,
We find that f(1)=2 is a minimum.
Since -1<=sin(x)<=1 for all real x, x cannot be real.
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