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  • 解方程2sin3x·cos2x+2sin^2x=0

    解方程2sin3x·cos2x+2sin^2x=0
    數(shù)學(xué)人氣:286 ℃時(shí)間:2020-04-08 08:55:03
    優(yōu)質(zhì)解答
    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<0
    ∴sinx=1==>x1=2kπ+π/2
    Sinx=0==>x2=2kπ,x3=2kπ+π
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