高中ko講義數學(一)
3. 若3sin2θ＋2cos2θ＝3，求tanθ之值。 解：設t＝tanθ，則 sin2θ＝＝，cos2θ＝＝ 故3sin2θ＋2cos2θ＝3化為 3×＋2×＝3 ( 5t2－6t＋1＝0 ( ( 5t－1 ) ( t－1 )＝0 解得t＝或1，即tanθ＝或1 4. 設tanα，tanβ為方程式x2－3x－2＝0的二根，求之值。 ................
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