Question 928167
Substitute,
{{{u=cos(t)}}}
Then,
{{{2u^2-u-1=0}}}
{{{(u-1)(2u+1)=0}}}
Two "u" solutions:
{{{u-1=0}}}
{{{u=1}}}
{{{cos(t)=1}}}
{{{t=0}}}
and
{{{2u+1=0}}}
{{{2u=-1}}}
{{{u=-1/2}}}
{{{cos(t)=-1/2}}}
{{{t=(2/3)pi}}} and {{{t=(4/3)pi}}}