MT【41】利用不等式妙消参数

已知$\theta\in[0,2\pi]$对任意$x\in[0,1],2x^2sin\theta-4x(1-x)cos\theta+3(1-x)^2>0$恒成立.求$\theta$的范围.

解答:令$x=1$易得$sin\theta>0,\because x\in(0,1)$,$$2x^2sin\theta-4x(1-x)cos\theta+3(1-x)^2$$

$$\ge2\sqrt{6}x(1-x)\sqrt{sin\theta}-4x(1-x)cos\theta$$ $$=2x(1-x)(\sqrt{6sin\theta}-2cos\theta)>0$$

$$\therefore 6sin\theta>4cos^2\theta$$ 或者$cos\theta\le0$易知:$$\theta\in(30^0,180^0)$$