急 ! F6 sin/cos數1題 #17b

(a) As sin θ is a root of the equation 3x^2 + kx + √3 = 0, so3 sin^2 θ + k sin θ + √3 = 0 ...... (i)cos θ is a root of the equation √3x^2 + kx + 3 = 0, so√3 cos^2 θ + k cos θ + 3 = 0 .... (ii)(i)*cos θ - (ii)*sin θ, we get,3 sin^2 θ cos θ + √3 cos θ - √3 sin θ cos^2 θ - 3 sin θ = 0==> sin θ cos θ (3 sin θ - √3 cos θ) - (3 sin θ - √3 cos θ) = 0==> (3 sin θ - √3 cos θ)(sin θ cos θ - 1) = 0==> 3 sin θ - √3 cos θ = 0 or sin θ cos θ - 1 = 0 (reject, as cos θ > 1/sin θ)==> tan θ = √3 / 3
(b) θ lies in Q3 as cos θ > 1/sin θ, therefore, sin θ = -1/2Sub. sin θ = -1/2 into (i), we get,3*(1/4) - k/2 + √3 = 0==> 3 - 2k + 4√3 = 0==> k = (3 + 4√3)/2

2014-03-07 20:23:30 補充：
貓Sir, if sin θ < 0, then
cos θ > 1/sin θ
==> sin θ cos θ < 1
不是大過 1 的。

yes, you are right.

2014-03-07 20:24:21 補充：
I realized this after reading your answer and before seeing your comments.