How do you find all solutions of the equations $2sec^2x+tan^2x-3=0$ in the interval $[0,2pi)$ ?

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Answer 1 · 2017-10-09T11:28:48

$"Solution set is " {pi/6,(5pi)/6,(7pi)/6,(11pi)/6}$

$2sec^2x+tan^2x-3=0$

we need to change to either all tangent or secant functions

now
$1+tan^2x=sec^2x$

so

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

$2+2tan^2x+tan^2x-3=0$

$=>3tan^2x=1$

$=>tanx=+-1/sqrt3$

we only need the positive square root since the interval specified is

$[0,2pi)$

and in $[0,2[i)$ , $x=pi/6, pi+pi/6=(7pi)/6$ or $x=(5pi)/6,(11pi)/6$

$"Solution set is " {pi/6,(5pi)/6,(7pi)/6,(11pi)/6}$

How do you find all solutions of the equations 2sec^2x+tan^2x-3=02sec^2x+tan^2x-3=0 in the interval [0,2pi)[0,2pi)?