已知3π/4

问题描述:

已知3π/4

tanx+1/tanx=-10/3 tanx=-1/3 tanx=-3
因为3π/4{5sin^2(x/2)+8sin(x/2)cos(x/2)+11cos^2(x/2)-8}/{√2sin(x-π/4)}
=(5(1-cosx)/2+4sinx+11(1+cosx)/2-8)/(sinx-cosx)
=(3cosx+4sinx)/(sinx-cosx)
=4+7cosx/(sinx-cosx)=4+1/{tanx/7-1/7}=4+1/((-1/3)/7-1/7)= -5/4

(13-根号10)/4

令tanx=t,那么1/tanx=1/t
t+1/t=-10/3
t×1/t=1
所以t和1/t是y²+10/3y+1=0的2个根
3y²+10y+3=0
(3y+1)(y+3)=0
y=-1/3或y=-3
因为3π/4