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How do you solve $4-3sec^2x=0$ ?

1 Answer

Nov 20, 2016

$4-3sec^2x=0$

$=>3sec^2x=4$

$=>sec^2x=4/3$

$=>cos^2x=3/4$

$=>cosx=+-sqrt3/2$

when

$cosx=+sqrt3/2=cos (pi/6)$

then $x=2npi+-pi/6." where " n in ZZ$

when

$cosx=-sqrt3/2=cos (pi-pi/6)=cos((5pi)/6)$

then $x=2npi+-(5pi)/6." where " n in ZZ$

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