Question 1025159
Using the trigonometric identities that we should all remember,
{{{tan(x)=sin(x)/cos(x)}}} and {{{sec(x)=1/cos(x)}}} ,
you can re-write the equation as
{{{sin^2(x)/cos^2(x)=1-1/cos(x)}}} .
Multiplying both sides of the equal sign times {{{cos^2(x)}}} ,
the equation above turns into
{{{sin^2(x)=cos^2(x)-cos(x)}}} ,
and since {{{sin^2(x)=1-cos^2(x)}}} ,
you can substitute and re-write it as
{{{1-cos^2(x)}=cos^2(x)-cos(x)}}} <--> {{{0=2cos^2(x)-cos(x)-1}}} .
The last equation is a quadratic equation.
If you use {{{y=cos(x)}}} , you can write it as
{{{2y^2-y-1=0}}}
You can solve for {{{y}}} using factoring, or completing the square, or the quadratic formula to get the solutions
{{{system(y=1,"or",y=-1/2)}}} .
In the interval {{{"[ 0 ,"}}}{{{2pi}}}{{{")"}}} ,
{{{cos(x)=1}}} happens only for {{{highlight(x=0)}}} ,
while {{{cos(x)=-1/2}}} happens only for {{{highlight(system(x=2pi/3,"or",x=4pi/3))}}} .