[
,
]
分析:設(shè)(a,b)為f(x)上任意一點(diǎn),設(shè)此點(diǎn)關(guān)于點(diǎn)
對(duì)稱的點(diǎn)為:(x,y),建立(a,b)與(x,y)的關(guān)系,求出g(x),最后求出x的范圍即可.
解答:∵f(x)=sinx,x∈R,而g(x)的圖象與f(x)的圖象關(guān)于點(diǎn)
對(duì)稱,設(shè)(a,b)為f(x)上任意一點(diǎn),
設(shè)此點(diǎn)關(guān)于點(diǎn)
對(duì)稱的點(diǎn)為:(x,y),根據(jù)題意有:
,解得
.
∵(a,b)為f(x)上任意一點(diǎn),∴b=sina,即:-y=sin(
-x),∴y=g(x)=-cosx.
∴在區(qū)間[0,2π]上,由f(x)≥g(x)可得sinx≤-cosx,即 sinx+cosx≤0,即
sin(x+
)≤0,即 sin(x+
)≤0.
故有 π≤x+
≤2π,由此可得x的范圍是:
≤x≤
,
故答案為[
,
].
點(diǎn)評(píng):本題主要考查正弦函數(shù)的對(duì)稱性,求一個(gè)函數(shù)圖象關(guān)于某個(gè)點(diǎn)對(duì)稱的圖象所對(duì)應(yīng)的函數(shù)解析式,屬于中檔題.