函数y=log以三分之一为底(sin2x-cos2x)的单减区间。

显示全部楼层 · 2012-8-3 22:59:19

解：y=log以三分之一为底(sin2x-cos2x)y=[lg(sin2x-cos2x)]/[lg(1/3)]y=-(lg3)lg(sin2x-cos2x)y'=-2[(lg3)/(ln10)][(cos2x+sin2x)/(sin2x-cos2x)]令：y'＜0有：-2[(lg3)/(ln10)][(cos2x+sin2x)/(sin2x-cos2x)]＜0即：(cos2x+sin2x)/(sin2x-cos2x)＞0(tan2x+1)/(tan2x-1)＞0有：tan2x＞-1、tan2x＞1………………(1)或：tan2x＜-1、tan2x＜1………………(2)由(1)得：tan2x＞1，即：kπ/2...

千问 · 2012-8-3 22:59:19

解：求导得：y'=(2/ln3)*(1+sin4x)/cos4x
令y'<0,得：x∈(π/8+kπ/2,3π/8+kπ/2),k∈Z
因此，原函数的单减区间为(π/8+kπ/2,3π/8+kπ/2),k∈Z。...

千问 · 2012-8-3 22:59:19

复合函数㏒1/3底t它是减函数∴t=sin2x-cos2x单调递增t=√2sin﹙2x－π/4﹚-π/2＋2kπ ≦2x－π/4≦ π/2＋2kπ-π/8＋kπ≦ x≦3π/8＋kπ k∈z...

千问 · 2012-8-3 22:59:19

解：求导得：y'=(2/ln3)*(1+sin4x)/cos4x令y'<0,得：x∈(π/8+kπ/2,3π/8+kπ/2),k∈Z因此，原函数的单减区间为(π/8+kπ/2,3π/8+kπ/2),k∈Z。...

千问 · 2012-8-3 22:59:19

(tan2x-1)＞0有：tan2x＞-1;2+π/[lg(1/＜0有;2+π/、tan2x＞1………………(1)或;4即：函数的单调减区间是：x∈(kπ/2-π/4，kπ/：kπ/2+π/8＞x＞kπ/2-π/=-2[(lg3)/：-2[(lg3)/(ln10)][(cos2x+sin2x)/(sin2x-cos2x)]＜0即：(cos...