How do you solve #2tan^2x=3tanx+7# in the interval [0,360]?

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Answer 1 · 2016-09-28T06:29:45

Solution is #{70.12^o,128.31^o,231.69^o,250.12^o}#

#2tan^2x=3tanx+7# can be written as

#2tan^2x-3tanx-7=0#

Using quadratic formula

#tanx=(-(-3)+-sqrt((-3)^2-4xx2xx(-7)))/(2xx2)#

= #(3+-sqrt(9+56))/4#

= #(3+-sqrt65)/4=(3+-8.06225)/4#

i.e. #tanx=11.06225/4=2.76556# or #tanx=-5.06226/4=-1.26556#

and using tables #x=tan^(-1)2.76556=70.12^o#

or #x=(180+70.12)^o=250.12^o#

as function tan has a cycle of #pi=180^o#

or #x=tan^(-1)(-1.26556)=(180-51.69)^o=128.31^o# or #x=(180+51.69)^o=231.69^o#

Hence Solution is #{70.12^o,128.31^o,231.69^o,250.12^o}#