Question 869656
{{{cos(2theta)=(cos(theta))^2-(sin(theta))^2=(cos(theta))^2+1-(sin(theta))^2-1=2(cos(theta))^2-1}}}
So,
{{{cos(2theta)+cos(theta)=0}}} can be re-written as
{{{2(cos(theta))^2+cos(theta)-1=0}}}
and if we abbreviate it by defining {{{y=cos(theta)}}}
we can re-write it as
{{{2y^2+y-1=0}}}
That quadratic equation can be solve by factoring,
or by completing the square,
or by using the quadratic formula,
to yield the solutions
{{{system(y=-1,"or",y=1/2)}}} .
Going back to {{{theta}}} and for {{{0^o<=theta<360^o}}}
we find that {{{y=cos(theta)=-1}}} ---> {{{highlight(theta=180^o)}}}
and {{{y=cos(theta)=1/2}}} ---> {{{system(highlight(theta=60^o),"or",highlight(theta=300^o))}}}