Question

How do you use the graphing utility to approximate the solution `tan(x+pi)-cos(x+pi/2)=0` in the interval `[0,2pi)`?

Answer 1

`kpi, k = 0, +-1, +-2, +-3, ...`. The graph cuts the x-axis at x = 0, +-3.1416, in one period `(-pi, pi)`, with vertical asymptotes `uarrx = +-pi/2darr`, therein.

Explanation:

As #tan(x+pi)=tanx and cos(x+pi/2)=-sinx, the equation simplifies to

`tan x+ sin x = sin x ( 1/cos x|1)=0`. giving

`sinx = 0= sin 0` that gives x = kpi, k = 0, +-1, +-2, +-3, ...#

and/or

`cos x=-1=cos pi that gives x = (2k|1)pi, k = 0, +-1, +-2, +-3, ...`

and this included in `kpi, k = 0, |=1, +-2, +-3, ...`

graph{tan x +sin x [-15.61, 15.6, -7.8, 7.81]}