Solve for x in radians sec2x = 2 ?

2 Answers
Apr 18, 2018

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

Explanation:

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}

Apr 18, 2018

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

Explanation:

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