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}#