作业帮解方程2sin3x·cos2x+2sin^2x=0
来源：学生作业帮 编辑：作业帮 分类：数学作业 时间：2024/05/11 17:22:31
解方程2sin3x·cos2x+2sin^2x=0
2*(sin(x))^3*cos(2*x)+2*(sin(x))^2=0
解析：左式= 2*(sin(x))^3*(1-2(sin(x))^2)+2*(sin(x))^2
= 2*(sin(x))^3-4*(sin(x))^5+2*(sin(x))^2
= [sin(x)-2*(sin(x))^3+1]2*(sin(x))^2
=(sin(x)-1)*(-2*(sin(x))^2-2*sin(x)-1)*2*(sin(x))^2=0
∵-2*(sin(x))^2-2*sin(x)-1x1=2kπ+π/2
Sinx=0==>x2=2kπ,x3=2kπ+π
解析：左式= 2*(sin(x))^3*(1-2(sin(x))^2)+2*(sin(x))^2
= 2*(sin(x))^3-4*(sin(x))^5+2*(sin(x))^2
= [sin(x)-2*(sin(x))^3+1]2*(sin(x))^2
=(sin(x)-1)*(-2*(sin(x))^2-2*sin(x)-1)*2*(sin(x))^2=0
∵-2*(sin(x))^2-2*sin(x)-1x1=2kπ+π/2
Sinx=0==>x2=2kπ,x3=2kπ+π
解方程2sin3x·cos2x+2sin^2x=0
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