解方程2sin3x·cos2x+2sin^2x=0
问题描述：
解方程2sin3x·cos2x+2sin^2x=0
答
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)-1∴sinx=1==>x1=2kπ+π/2
Sinx=0==>x2=2kπ,x3=2kπ+π