Question

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

Answer 1

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

Explanation:

`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}`