【1.y=sin(1/2x+π/6),x属于[0,π/3]2.y=-cos(3x-π/3),x属于[-π/3,π/3]】
<p>问题：【1.y=sin(1/2x+π/6),x属于2.y=-cos(3x-π/3),x属于[-π/3,π/3]】<p>答案：↓↓↓<p class="nav-title mt10" style="border-top:1px solid #ccc;padding-top: 10px;">罗国涛的回答：<div class="content-b">网友采纳 1 y=sin(1/2x+π/6) x属于 π/6≤1/2x+π/6≤π/3 1/2≤sin(1/2x+π/6)≤根号3/2 值域【1/2,根号3/2】 2 y=-cos(3x-π/3) x属于[-π/3,π/3] -π-π/3≤3x-π/3≤π-π/3 (3x-π/3)的变化范围涵盖了从--π-π/3到π-π/3正好2π的区间 ∴y=-cos(3x-π/3)值域为【-1,1】
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