Question

已知sinx＋cosx=-1/5（0＜x＜π），求tanx的ŀ

Answer 1

sinx＋cosx=-1/5（0＜x＜π）①
所以3π/4<x<π。设t=tanx>-1,
①平方得1+2sinxcosx=1/25,
所以2t/(1+t^2)=-24/25,
25t=-12-12t^2,
整理得12t^2+25t+12=0,
t=(-25土7）/24=-3/4，或-4/3(舍).

Answer 2

0＜x＜π 时，sinx＋cosx = √2sin(x+π/4) = -1/5,
π < x+π/4 < 3π/2, 3π/4 < x < 5π/4
sinx＋cosx = cosx(1+tanx) = (1+tanx)/secx

= (1+tanx)/{-√[1+(tanx)^2]} = -1/5
(1+tanx)^2/[1+(tanx)^2] = 1/25
25+50tanx+25(tanx)^2 = 1+(tanx)^2
12(tanx)^2+25tanx+12 = 0
(4tanx+3)(3tanx+4) = 0
tanx = -3/4, 或 tanx = -4/3 （π/2 < x < 3π/4， 故舍弃） ，
tanx = -3/4

Answer 3

sinX=3/5.
cosX=-4/5.
0<X<180"
X=150"
tanX=-3/4.