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一些求参数取值范围的问题可以转化为求最值的问题例1 当a取何实数时,方程2acos~2x-sinx+2+a=0有实数解? 解:由原方程解出a=(sinx+2)/(2cos~2x+1)=(sinx-2)/(3-2sin~2x)∴1/a=(2sin~3x-3)/(2-sinx)=-2sinx-4+5/(2-sinx) 设t=2-sinx∈[1,3]。化1/a=2t+5/t-8=(2t~(1/2)-(5/t)~(1/2)+2(10)~(1/2)-8 故在(2t)~(1/2)=(5/t)~(1/2)即t=5~(1/2)/2~(1/2)=2-sinx 即sinx=4-(10)~(1/2)/2(∈[-1,1])时1/a取最小值2(10)~(1/2)-8
The problem of finding the range of parameters can be converted into the problem of the best value. Example 1 When a is a real number, the equation 2acos~2x-sinx+2+a=0 has a real solution. Solution: It is solved by the original equation a= (sinx+2)/(2cos~2x+1)=(sinx-2)/(3-2sin~2x)∴1/a=(2sin~3x-3)/(2-sinx)=-2sinx-4 +5/(2-sinx) Let t=2-sinx∈[1,3]. 1/a=2t+5/t-8=(2t~(1/2)-(5/t)~(1/2)+2(10)~(1/2)-8 )~(1/2)=(5/t)~(1/2) ie t=5~(1/2)/2~(1/2)=2-sinx ie sinx=4-(10)~ (1/2)/2(∈[-1,1]) When 1/a takes the minimum value 2(10)~(1/2)-8