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 ${70.12^o,128.31^o,231.69^o,250.12^o}$

How do you solve 2tan^2x=3tanx+72tan^2x=3tanx+7 in the interval [0,360]?