Question 932052
Use a substitution,
{{{cos^2(theta)=1-sin^2(theta)}}}.
Then,
{{{1-sin^2(theta)=sin(theta)-.02}}}
{{{sin^2(theta)+sin(theta)-1.02=0}}}
Now use another substitution,
{{{u=sin(theta)}}}
{{{u^2+u-1.02=0}}}
Quadratic equation, you can solve by quadratic formula or completing the square.
{{{u^2+u+1/4-1.02=1/4}}}
{{{(u+1/2)^2=1/4+102/100}}}
{{{(u+1/2)^2=25/100+102/100}}}
{{{(u+1/2)^2=127/100}}}
{{{u+1/2=0 +- sqrt(127)/10}}}
{{{u=-1/2 +- sqrt(127)/10}}}
So then,
{{{sin(theta)=-1/2 +- sqrt(127)/10}}}
Since {{{abs(sin(theta))<1}}}, then the value, {{{-1/2-sqrt(127)/10}}} is not allowed since it is outside of this range.
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{{{sin(theta)=sqrt(127)/10-1/2}}}
{{{sin(theta)=0.6269}}}
{{{theta=38.8}}} and {{{theta=141.2}}}.
Both are in units of degrees.