Question 162318
This one looks like a trig relation to me. 

Remember that
{{{cos^2(theta)+sin^2(theta)=1}}}.
This will rearrange easily to 
{{{cos^2(theta)=1-sin^2(theta)}}}.

Put that on the right hand side of the initial equation you gave me:
{{{sin^2(theta)-sin(theta)=cos^2(theta)}}}
with our new right hand side:
{{{sin^2(theta)-sin(theta)=1-sin^2(theta)}}}.

Now we can rearrange to get zero on the right hand side:
{{{2*sin^2(theta)-sin(theta)-1=0)}}}.

We treat this like a quadratic equation of the form {{{2x^2 -x -1}}} where x=sin(theta)

The equation factorises to:
{{{(2*sin(theta)+1/2)(sin(theta)-1)=0)}}}.

So
{{{2*sin(theta)+1/2=0}}}
or
{{{sin(theta)-1)=0}}}

keep the sin(theta on the left hand sides of these and rearrange to get
{{{sin(theta)=-1/4}}}
or
{{{sin(theta)=1}}}

{{{sin^-1(-1/4)}}}=-4.6083,184.6083
and
{{{sin^-1(1)}}}=0,180,360

convert to radians by multiplying by {{{180/pi}}}


Hope this helps,
Adam