Question 868732
Use the trigonometric identity {{{cos(2x)=(cos(x))^2-(sin(x))^2}}} .
Since {{{(cos(x))^2=1-(sin(x))^2}}} , it can be re-arranged to
{{{cos(2x)=1-2(sin(x))^2}}} .
Substituting into {{{ cos(2x)+sin(x)=1}}} you get
{{{1-2(sin(x))^2+sin(x)=1}}}<--->{{{-2(sin(x))^2+sin(x)=0}}} (subtracting {{{1}}} from both sides of the equal sign)
That is a quadratic equation in {{{sin(x)}}} .
It looks complicated, but if you rename {{{y=sin(x)}}} ,
you can re-write it and it looks very simple:
{{{-2y^2+y=0}}}<-->{{{2y^2-y-0}}}<-->{{{y(2y-1)=0}}}-->{{{system(y=0,"or",y=1/2)}}}
So, {{{-2(sin(x))^2+sin(x)=0}}} is true when
{{{system(sin(x)=0,"or",sin(x)=1/2)}}} .
In {{{"[ 0 ,"}}}{{{2pi}}}{{{")"}}} ,
{{{sin(x)=0}}} --> {{{system(highlight(x=0),"or",highlight(x=pi))}}} and
{{{sin(x)=1/2}}} --> {{{system(highlight(x=pi/6),"or",highlight(x=5pi/6))}}}