How do i solve for the value of x? sec^2(2x)-4=0

Question

[0,2pi)

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Answer 1 · 2018-03-02T09:36:53

#color(blue)(x=pi/6 , (5pi)/6, pi/3 , (2pi)/3)#

#color(red)bb(sec^2(2x)=1/cos^2(2x))#

#1/cos^2(2x)-4=0#

#(1-4cos^2(2x))/cos^2(2x)=0#

#1-4cos^2(2x)=0#

#cos^2(2x)=1/4#

#cos(2x)=+-1/sqrt(4)=1/2,-1/2#

#2x=arccos(cos(2x))=arccos(1/2)=>2x=pi/3,(5pi)/3#

#x=pi/6 , (5pi)/6#

#2x=arccos(cos(2x))=arccos(-1/2)=>2x=(2pi)/3,(4pi)/3#

#x=pi/3 , (2pi)/3#

All solutions:

#color(blue)(x=pi/6 , (5pi)/6, pi/3 , (2pi)/3)#