t∈[π/6,2π+π/6]是t的取值范围
而t∈[π/6,π/2]或t∈[3π/2,2π+π/6]
是函数的单调增区间
∵sinα的单调递增区间为[-π/2+2kπ，π/2+2kπ]
∴k取0时，为[-π/2，π/2]得出t∈[π/6,π/2]
k取1时，为[3π/2，5π/2]得出t∈[3π/2,2π+π/6]
∴可得出t∈[π/6,π/2]或t∈[3π/2,2π+π/6]
只帮你理解答案，不从解题的角度，你先不要管X，只考虑t，f(t)=2sin(t)+1的单调递增区间为t∈[π/6,π/2]或t∈[3π/2,2π+π/6]，这个好理解吧？然后在考虑t=2x+π/6，t∈[π/6,π/2]或t∈[3π/2,2π+π/6]时，x∈[？]就行了
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