
已知函数f(x)=cos²﹙x+π/12﹚,g﹙x﹚=1+1/2sin2x,非常急加分 10
令h(x)=f(x)+g(x),若T(x)为偶函数且周期为5TT/3,当x属于(0,5TT/6)时,T(x)=2h(x)-3,求方程T(x)=1/2在区间(0,2TT)上...
令h(x)=f(x)+g(x),若T(x)为偶函数且周期为5TT/3,当x属于(0,5TT/6)时,T(x)=2h(x)-3,求方程T(x)=1/2在区间(0,2TT)上的解集
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h(x)=f(x)+g(x)=cos²﹙x+π/12﹚+1+1/2sin2x
=½+½cos﹙2x+π/6﹚+1+½sin2x
=½+½[cos﹙2x﹚cosπ/6-sin2xsinπ/6]+1+½sin2x
=½[cos﹙2x﹚cosπ/6+sin2xsinπ/6]+3/2
=½cos﹙2x-π/6﹚+3/2
T(x)=2h(x)-3=cos﹙2x-π/6﹚,x∈﹙0,5π/6﹚
画图可得2x1-π/6=π/3,即x1=π/4,x2=﹣π/4+5π/3=17π/12,
x3=π/4+5π/3=23π/12
解集﹛π/4,17π/12,23π/12﹜
=½+½cos﹙2x+π/6﹚+1+½sin2x
=½+½[cos﹙2x﹚cosπ/6-sin2xsinπ/6]+1+½sin2x
=½[cos﹙2x﹚cosπ/6+sin2xsinπ/6]+3/2
=½cos﹙2x-π/6﹚+3/2
T(x)=2h(x)-3=cos﹙2x-π/6﹚,x∈﹙0,5π/6﹚
画图可得2x1-π/6=π/3,即x1=π/4,x2=﹣π/4+5π/3=17π/12,
x3=π/4+5π/3=23π/12
解集﹛π/4,17π/12,23π/12﹜
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那题目当中给的,若T(x)为偶函数且周期为5TT/3怎么用呢
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画图:先根据T(x)为偶函画出[﹣5π/6,0]的图像,再根据周期为5π/3画出[5π/6,2π﹚的图像
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