Question 358876
{{{ (sinx)^2 - cosx+3=0}}},
{{{1-(cosx)^2 - cosx+3=0}}},after using the Pythagorean identity.
{{{-(cosx)^2 -cosx+4=0}}},
{{{(cosx)^2 +cosx-4=0}}}.
Applying the quadratic formula,
{{{cosx = (-1 +- sqrt( 1^2-4*1*-4 ))/(2*1)}}},
{{{cosx = (-1+-sqrt(17))/2}}}.
However, these values are greater than 1, and also less than -1.  Therefore the trigonometric equation does not have roots.