How do you solve #tan^2x-2sec x+1=0# for #0°<=x<=360°# ?

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Answer 1 · 2015-09-29T00:59:43

#x=60^@,x=300^@#

As #tan^2(x)=sec^2(x)-1# we can replace the #tan^2(x)# in the initial equation.

This leaves us with;

#sec^2(x)-1-2sec(x)+1=0#

After combining like terms the resulting formulae is:

#sec^2(x)-2sec(x)=0#

We can then factorise by #sec(x)# to yield;

#sec(x)(sec(x)-2)=0#

By the null factor law we can then establish two possible solutions for x: when #sec(x)=0# or when #sec(x)=2#

As the secant of x can never equal zero this possibility can be discounted which leaves #sec(x)=2#.

To solve for #sec(x)=2# first we take the reciprocal of each side;

#cos(x)=1/2#

Now the values of x that satisfy this equality that are within the domain #[0^@,360^@]# are; #x=60^@,x=300^@#.

Hope this helps :)