1.(1) ∵y=tan(π/根號(hào)3 ×sinx)
∴(π/根號(hào)3 ×sinx)∈(-π/2+kπ,π/2+kπ)
∴-根號(hào)3/2＜sinx＜根號(hào)3/2 (-1＜sinx＜1)
∴函數(shù)定義域?yàn)?kπ,kπ+π/3)∪(kπ+2π/3,kπ+2π)
∵-π/根號(hào)3＜π/根號(hào)3 ×sinx＜π/根號(hào)3
∴函數(shù)值域?yàn)椋?-tan π/根號(hào)3,tan π/根號(hào)3) (這個(gè)如果你們老師要求算出確切值的話用計(jì)算器算就好了.)
(2)∵y=tanx在(-/2,π/2)上單調(diào)遞增
∴當(dāng)x∈(-π/2,π/2)時(shí),sinx單調(diào)遞增,f(x)單調(diào)遞增
當(dāng)x∈(-π,-π/2)∪(π/2,π)時(shí),sinx單調(diào)遞減,f(x)單調(diào)遞減.
(3)∵f(x)=tan根號(hào)2/3π
∴T=3π²/2
∴根據(jù)圖像得：函數(shù)在(-π,π)上的解的個(gè)數(shù)為1