Question 174834
Let {{{a}}} = sin theta
{{{3a^2 - 4a + 1 = 0}}}
{{{a = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}
{{{a = (-(-4) +- sqrt( (-4)^2-4*3*1 ))/(2*3) }}}
{{{a = (4 +- sqrt(16 - 12 ))/6 }}}
{{{a = (4 +- sqrt(4))/6 }}}
{{{a = 1}}}
{{{a = 1/3}}}
sin theta = 1
(1) theta = 90 degrees
sin theta = 1/3
(2) theta = 19 degrees 28 min
(3) theta = 199 degrees 28 min