Question 1026450
This is a quadratic equation.
Let {{{u=sin(theta)}}}
{{{u^2-7u-8=0}}}
{{{(u-8)(u+1)=0}}}
Two "u" solutions:
{{{u-8=0}}}
{{{u=8}}}
and
{{{u+1=0}}}
{{{u=-1}}}
Back substituting,
{{{sin(theta)=8}}}
There is no theta solution here since sin can never exceed 1.
.
.
{{{sin(theta)=-1}}}
{{{theta=(3/2)pi+2pi*k}}} where  k is any integer.