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}#