Question 747135
{{{(cos(x))^2+(sin(x))^2=1}}} --> {{{(sin(x))^2=1-(cos(x))^2}}} so
{{{(sin(x))^2+cos(x)+1=0}}} --> {{{1-(cos(x))^2+cos(x)+1=0}}} --> {{{-(cos(x))^2+cos(x)+2=0}}} --> {{{(cos(x))^2-cos(x)-2=0}}}
Calling {{{cos(x)=y}}} we can re-write the equation as
{{{y^2-y-2=0}}} --> {{{(y+1)(y-2)=0}}}
The solutions to that equation are {{{y=2}}} and {{{y=-1}}},
but since {{{-1<=cos(x)<=1}}}, {{{y=2}}} does not yield a solution of the original
equation.
Looking for solutions between {{{0^o}}} and {{{360^o}}} (between {{{0}}} and {{{2pi}}}radians),
{{{y=cos(x)=-1}}} --> {{{x=180^o}}} (or {{{x=pi}}} radians)
All solutions can be written as
{{{x=(2k+1)*180^o}}} (or {{{x=(2k+1)*pi}}} if measuring angles in radians)