Solve for x in radians sec2x = 2 ?

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Answer 1 · 2018-04-18T16:24:10

#x in {1/6pi+kpi:k in ZZ}# or #x in{ 5/6pi+kpi: k in ZZ}#

#sec2x=2#

so #cos2x=1/2#

so #2x=arccos(1/2)+2kpi#

so #2x=1/3pi+2kpi# or #2x=5/3pi+2kpi# where #kinZZ#

so #x=1/6pi+kpi# or #x=5/6pi+kpi# where #k in ZZ#

so #x in {1/6pi+kpi:k in ZZ}# or #x in{ 5/6pi+kpi: k in ZZ}#

Answer 2 · 2018-04-18T16:34:22

#=>x=2kpi+-pi/6,kinZZ orx=2kpi+-(5pi)/6,kinZZ#

Weknow that,

#color(red)((1)cos2theta=2cos^2theta-1#

#color(blue)((2)costheta=cosalpha=>x=2kpi+-alpha,kinZZ#

Here,

#sec2x=2#

#=>color(red)(cos2x)=1/2...toApplycolor(red)((1)#

#=>color(red)(2cos^2x-1)=1/2#

#=>2cos^2x=3/2#

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

#=>cosx=+-sqrt3/2#

#=>cosx=sqrt3/2 or cosx=-sqrt3/2#

#=>cosx=cos(pi/6) or cosx=cos(pi-pi/6)=cos((5pi)/6)#

Using #(2)#, we get

#=>color(blue)(x=2kpi+-pi/6,kinZZ orx=2kpi+-(5pi)/6,kinZZ#