How do you solve #tanx(tanx-1)=0#?

1 Answer
Aug 4, 2016

Assuming a domain of #[0,2pi]#
#color(white)("XXX")color(green)(x in {0,pi/4,pi,(5pi)/4}#

Explanation:

Given (at least this is what I think the expression meant):
#color(white)("XXX")color(red)(""(tan(x)))*color(blue)(""(tan(x)-1))=0#

Either
#color(white)("XXX")color(red)(tan(x)=0)color(white)("XX")orcolor(white)("XX")color(blue)(tan(x)-1=0)#

#rArrcolor(white)("X")color(red)(x= 0" or " pi)color(white)("XXXXXX")color(blue)(tan(x)=1)#

#color(white)("XXXXXXXXXXXXXXXXX")color(blue)(rArrx=pi/4" or " (5pi)/4)#