Solve the equation $6sec^2x+3tan^2x-9=0$ in the interval $[0,2pi]$ ?

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Answer 1 · 2017-03-23T17:17:46

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

In $6sec^2x+3tan^2x-9=0$ putting $sec^2x=1+tan^2x$ , we get

$6(1+tan^2x)+3tan^2x-9=0$

or $9tan^2x-3=0$

or $3tan^2x-1=0$

or $(sqrt3tanx+1)(sqrt3tanx-1)=0$

i.e. $tanx=-1/sqrt3$ or $1/sqrt3$

and in the interval $[0,2pi]$

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

Solve the equation 6sec^2x+3tan^2x-9=06sec^2x+3tan^2x-9=0 in the interval [0,2pi][0,2pi]?