(1)解方程3sin2x-8(sinx)^2-1=0(2)求y=arcsin(1-x)+arccos2x的值域解方程3sin2x-8(sinx)^2-1=0 求y=arcsin(1-x)+arccos2x的值域
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![(1)解方程3sin2x-8(sinx)^2-1=0(2)求y=arcsin(1-x)+arccos2x的值域解方程3sin2x-8(sinx)^2-1=0 求y=arcsin(1-x)+arccos2x的值域](/uploads/image/z/52319-47-9.jpg?t=%EF%BC%881%EF%BC%89%E8%A7%A3%E6%96%B9%E7%A8%8B3sin2x-8%28sinx%29%5E2-1%3D0%EF%BC%882%EF%BC%89%E6%B1%82y%3Darcsin%281-x%29%2Barccos2x%E7%9A%84%E5%80%BC%E5%9F%9F%E8%A7%A3%E6%96%B9%E7%A8%8B3sin2x-8%28sinx%29%5E2-1%3D0+%E6%B1%82y%3Darcsin%281-x%29%2Barccos2x%E7%9A%84%E5%80%BC%E5%9F%9F)
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(1)解方程3sin2x-8(sinx)^2-1=0(2)求y=arcsin(1-x)+arccos2x的值域解方程3sin2x-8(sinx)^2-1=0 求y=arcsin(1-x)+arccos2x的值域
(1)解方程3sin2x-8(sinx)^2-1=0(2)求y=arcsin(1-x)+arccos2x的值域
解方程3sin2x-8(sinx)^2-1=0
求y=arcsin(1-x)+arccos2x的值域
(1)解方程3sin2x-8(sinx)^2-1=0(2)求y=arcsin(1-x)+arccos2x的值域解方程3sin2x-8(sinx)^2-1=0 求y=arcsin(1-x)+arccos2x的值域
(1) 3sin2x-8(sinx)^2-1=0 3sin2x-4(1-cos2x)-1=0 (3/5)sin2x+(4/5)cos2x=1 sin[2x+arcsin(4/5)]=1 2x+arcsin(4/5)=2kπ+π/2 (k∈Z) x=kπ+π/4-(1/2)arcsin(4/5).(2) 易求得定义域是:[0,1/2] 又这个函数在[0,1/2]上是减函数,所以x=0时y最大为π/2+π/2=π,x=1/2时y最小是:π/6+0=π/6 所以值域是[π/6,π].