f（x）是定义在R上的以3为周期的奇函数，且f（2）=0，则方程f（x）=0在区间（0，6）内解的个数的最小值是______．
由函数的周期为3可得f（x+3）=f（x）
由于f（2）=0，
若x∈（0，6），则可得出f（5）=f（2）=0，
又根据f（x）为奇函数，则f（-2）=-f（2）=0，
又可得出f（4）=f（1）=f（-2）=0，
又函数f（x）是定义在R上的奇函数，可得出f（0）=0，
从而f（3）=f（0）=0，在f（x+3）=f（x）中，
令x=-
，得出f(-3 2
)=f(3 2
)，3 2
又根据f（x）是定义在R上的奇函数，得出f(-
)=-f(3 2
)，3 2
从而得到f(
)=-f(3 2
)，即f(3 2
)=0，3 2
故f(
)=f(9 2
+3)=f(3 2
)=0，3 2
从而f(
)=f(9 2
)=f（4）=f（1）=f（3）=f（5）=f（2）=0，共7个解3 2
故答案为：7
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