How do you solve #sec(x/2)=cos(x/2)# over the interval 0 to 2pi?

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Answer 1 · 2016-02-23T22:05:38

#x=0# and #x=2pi#

First, use the identity that #sec(x)=1/cos(x)# to rewrite the equation:

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

Cross multiply, or multiply both sides by #cos(x/2)#. This has the same function:

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

Take the square root of both sides.

#cos(x/2)=+-1#

Thus, we see that #x/2=0# and #x/2=pi#, since #cos(0)=1# and #cos(pi)=-1#.

#x/2=0" "=>" "x=0#

#x/2=pi" "=>" "x=2pi#

While there are an infinite number of solutions, these are the only two on the interval.