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# Express as a quadratic function of cosθ watch

1. Express 2sin2θ + 3cosθ as a quadratic function of cosθ.

Can somebody explain how I would go about doing this?
The answer sheet says that the answer is 2 - 2cos2θ + 3cosθ
2. (Original post by jamesthelam)
Express 2sin2θ + 3cosθ as a quadratic function of cosθ.

Can somebody explain how I would go about doing this?
The answer sheet says that the answer is 2 - 2cos2θ + 3cosθ
use the trig identiy sin2x+cos2x = 1
rearrange and sub in and youll get 2(1-cos2x)+3cosx=0
rearrange again and youll get that answer
3. (Original post by jamesthelam)
Express 2sin2θ + 3cosθ as a quadratic function of cosθ.

Can somebody explain how I would go about doing this?
The answer sheet says that the answer is 2 - 2cos2θ + 3cosθ
As the answer is in terms of Cos, I suspect that you have to change sinsquared into terms of cos. I think there's a standard identity for that.
4. (Original post by Daniel9998)
use the trig identiy sin2x+cos2x = 1
rearrange and sub in and youll get 2(1-cos2x)+3cosx=0
rearrange again and youll get that answer

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