How do you find the exact value of $3tan^2theta-4tantheta-1=0$ in the interval $0<=theta<2pi$ ?

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Answer 1 · 2016-12-08T15:20:52

The answer is $={57.1º,237.1º,347.9º,167.9º}$

You solve this like a quadratic equation

$ax^2+bx+c=0$
Where you replce $x$ by $tantheta$

We calculate, first the determinant

$Delta=b^2-4ac=(-4)^2-4(3)(-1)$

$=16+12=28$

As, $Delta>0$ , we have 2 real roots

$x=(-b+-sqrtDelta)/(2a)$

So,

$tantheta=(4+-sqrt28)/6$

$=(2+-sqrt7)/3$

$theta_1=arctan((2-sqrt7)/3)=347.9$ º and $=167.9º$

and $theta_2=arctan((2+sqrt7)/3)=57.1$ º and $=237.1º$

How do you find the exact value of 3tan^2theta-4tantheta-1=03tan^2theta-4tantheta-1=0 in the interval 0<=theta<2pi0<=theta<2pi?