Question

若2sinθ+3cosθ=0，求2cos2θ+3sin2θ的值

Answer 1

2sinθ+3cosθ=0
sinθ=-3cosθ/2
代入恒等式sin²θ+²θ=1
所以(9/4+1)cos²=1
cos²θ=4/13
sin²θ=9/13

cos2θ=cos²θ-sin²θ=5/13

sin2θ=2sinθcosθ
=2(-3cosθ/2)cosθ
=-3cos²θ
=-12/13

所以原式=10/13-36/13=-2

Answer 2

由2sinθ+3cosθ=0 ,sinθ^2+cosθ^2=1列方程组得：
sinθ^2=9/13 cosθ^2=4/13 sinθ*cosθ=+-6/13
原式=4*cosθ^2-2+-6sinθcosθ=-14/13 或 -2

Answer 3

2sinθ+3cosθ=0
2sinθ=-3cosθ
sinθ/cosθ=-3/2
tanθ=-3/2

2cos2θ+3sin2θ
=2*[1-(tanθ)^2]/[1+(tanθ)^2]+3*2(tanθ)/[1+(tanθ)^2]
=2*(1-9/4)/(1+9/4)+6*(-3/2)/(1+9/4)
=(-5/2)/(13/4)+(-9)/(13/4)
=(-5/2-9)/(13/4)
=(-23/2)/(13/4)
=-23/2*4/13
=-46/13