msww.net
当前位置:首页 >> 已知sinxsin2xsin3x+CosxCos2xCos3x=1 求x的值 >>

已知sinxsin2xsin3x+CosxCos2xCos3x=1 求x的值

cos2xcos3x-sin2xsin3x=0 cos(2x+3x)=0 5x=kπ+π/2(k为整数) 所以x=π/lO或3π/lO或5π/lO…

sin2xsin3x=cos2xcos3x所以cos2xcos3x-sin2xsin3x=0即cos5x=05x=k*pi+pi/2k属于整数x=k*pi/5+pi/10所以选a

f(x)=sin2xsin3x+cos2xcos3x-1=cos(3x-2x)-1=cosx-1;当x=(2n+1)π时,有f(x)最小值为-2

1.sin2xsin3x=cos2xcos3x--> cos2xcos3x-sin2xsin3x=0--> cos(2x+3x)=0--> cos5x=0故5x=kπ+π/2(k∈Z)x=kπ/5+π/10(k∈Z)k=0时x=π/10,故x的一个值为π/10.2.设t=cosAcosB.1/2+t=sinAsinB+cosAcosB=cos(A-B)t-1/2=cosAcosB-sinAsinB=cos(A+B)故-1≤1/2+t≤1,-1≤t-1/2≤1解得-1/2≤t≤1/2,即cosAcosB的取值范围是[-1/2,1/2].

设sinx+cosx=tsinxcosx=[(sinx+cosx)-1]/2=(t-1)/2sinx+cosx=(sinx+cosx)(sinx+sinxcosx+cosx)=t[1+(t-1)/2]=1即t+t-2=0(t-1)(t+t+2)=0易知t+t+2>0故t=1即sinx+cosx=1故sinxcosx=0sin^4x+cos^4x=(sinx+cosx)-2sinxcosx=1

sinxsin3x=-(cos4x-cos2x)/2∫sinxsin2xsin3xdx=(1/4)∫sin4xdx - (1/2)∫sin2xcos4xdx=-(1/16)cos4x - (1/4)∫cos4xd(cos2x)=-(1/16)cos4x - (1/4)∫(2(cos2x)^2-1)d(cos2x)=-(1/16)cos4x + (1/4)cos2x - (1/2)∫(cos2x)^2d(cos2x)=-(1/16)cos4x + (1/4)cos2x - (1/6)(cos2x)^3

原式即化为cos2xcos3x-sin2xsin3x=0cos(2x+3x)=0所以5x=kπ+π/2用公式:cos(a+b)=cosacosb-sinasinb

利用三角函数的积化和差公式即可

sinx-cosx=1/3(sinx-cosx)^2=1/92sinxcosx=8/9sin2x=8/9sin^3 x - cos ^3 x=(sinx-cosx)(sin^2x+sinxcosx+cos^2x)=1/3 * (1+4/9)=13/27

解由cosacosb+sinasinb=cos(a-b)故y=cos2xcos3x+sin2xsin3x=cos(2x-3x)=cos(-x)=cosx

相关文档
网站首页 | 网站地图
All rights reserved Powered by www.msww.net
copyright ©right 2010-2021。
内容来自网络,如有侵犯请联系客服。zhit325@qq.com