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函数y=2sin(x+π/3),x属于[π/6,π/2]的值域
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解:
π/6<=x<=π/2
π/2<=x+π/3<=5π/6
1/2<=sin(x+π/3)<=1
1<=2sin(x+π/3)<=2
所以函数值域为[1,2]
π/6<=x<=π/2
π/2<=x+π/3<=5π/6
1/2<=sin(x+π/3)<=1
1<=2sin(x+π/3)<=2
所以函数值域为[1,2]
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π/6≤x≤π/2
π/2≤x+π/3≤5π/6
1/2≤sin(x+π/3)≤1≤
1≤2sin(x+π/3)≤2
π/2≤x+π/3≤5π/6
1/2≤sin(x+π/3)≤1≤
1≤2sin(x+π/3)≤2
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先画y=sinX的图像,再横坐标不变,纵坐标变为原来的2倍,再向左平移π/3个单位,然后找x属于[π/6,π/2]的值域
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