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已知2cosxcos(x-π/6)-根号三sin^2x+sinxcosx求f(x)的最小正周期 当x属于【0,π】时,若f(x)=1
求x的值
人气:196 ℃ 时间:2019-10-29 10:46:53
解答
f(x)=2cosxcos(x-π/6)-√3sin^2x+sinxcosx =2cosxcos(x-π/6)-√3sin^2x+sinxcosx=2cosx(√3/2cosx+1/2sinx)-√3in^2x+sinxcosx=√3cos^2x+sinxcosx-√3sin^2x+sinxcosx=√3(cos^2x-sin^2x)+2sinxcosx=√3cos2x+sin...
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