Question 870543
Let {{{u=cos(theta)}}}.
{{{2u^2+u-1=0}}}
{{{(u+1)(2u-1)=0}}}
Two solutions:
{{{u+1=0}}}
{{{u=-1}}}
{{{ cos(theta)=-1}}}
{{{theta=pi}}}
and
{{{2u-1=0}}}
{{{2u=1}}}
{{{u=1/2}}}
{{{cos(theta)=1/2}}}
{{{theta=pi/3}}} and {{{theta=(5/3)pi}}}