Solve the trigonometric equation #tan^2x+tanx-1=0# in general and in the interval #[0,2pi)#?

Question

#tan^2x+tanx-1=0#

5080 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-03-26T16:44:10

#x=nxx180^@+31.7^@# or
#x=nxx180^@-58.3^@#,
where #n# is an integer
and in #{0,2pi)#, #x={31.7^@,121.7^@,211.7^@,301.7^@}#

#tan^2x+tanx-1=0#

#:.tanx=(-1+-sqrt(1^2-4xx1xx(-1)))/2#

#=(-1+-sqrt5)/2=(-1+-2.236)/2#

i.e. #-1.618# or #0.618#

i.e. #x=-58.3^@#, #x=31.7^@#

Hence #x=nxx180^@+31.7^@# or #x=nxx180^@-58.3^@#, where #n# is an integer

and in #{0,2pi)#, #x={31.7^@,121.7^@,211.7^@,301.7^@}#