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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