How do you solve $(3tan^2x-1)(tan^2x-3)=0$ in the interval 0 to 2pi?

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Answer 1 · 2016-11-16T03:38:44

$x = {pi/6,(7pi)/6,(5pi)/6,(11pi)/6,pi/3,(4pi)/3,(2pi)/3,(5pi)/3}$

$(3tan^2x-1)(tan^2x-3)=0$
Either
$(3tan^2x-1)=0 or (tan^2x-3)=0$

$tanx=+-1/sqrt3$

when

$tanx=1/sqrt3=tan(pi/6)=tan(pi+pi/6)$

So $x= pi/6 or (7pi)/6$

when

$tanx=-1/sqrt3=tan(pi-pi/6)=tan(2pi-pi/6)$

So $x= (5pi)/6 or (11pi)/6$

From $(tan^2x-3)=0$ we get

$tanx=+-sqrt3$

when
$tanx=+sqrt3=tan(pi/3)=tan(pi+pi/3)$

So $x=pi/3 or (4pi)/3$

Again when

$tanx=-sqrt3=tan(pi-pi/3)=tan(2pi-pi/3)$

So $x=(2pi)/3 or (5pi)/3$

How do you solve (3tan^2x-1)(tan^2x-3)=0(3tan^2x-1)(tan^2x-3)=0 in the interval 0 to 2pi?