Question 72889
use the identity {{{(cos(x))^2=1-(sin(x))^2}}} and replace {{{(cos(x))^2}}}
{{{1-(sin(theta))^2=4sin(theta)+4}}}
Now let {{{x=sin(theta)}}}
{{{1-x^2=4x+4}}}
Get x to one side
{{{x^2+4x+3=0}}}
Solve for x
{{{(x+3)(x+1)=0}}}
Solutions are {{{x=-3}}} and {{{x=-1}}}
This means {{{sin(theta)=-3}}} and {{{sin(theta)=-1}}}
Since the first solution doesn't make any sense, ignore it. Solve for {{{sin(theta)=-1}}}
{{{arcsin(sin(theta))=arcsin(-1)}}}
{{{theta=-pi/2+pi*n}}}Since {{{-pi/2}}} is not in the interval 0<(theta)<2pi it can be ignored also
Now find the other value
{{{theta=pi-arcsin(-1)}}}This allows us to go the other value of sine 
{{{theta=pi-(-pi/2)}}}
{{{theta=(2pi+pi)/2}}}
{{{theta=3pi/2}}}There's our other value
So {{{theta=3pi/2}}} is our answer.
<p>
Check:
{{{(cos(3pi/2))^2 = 4sin(3pi/2)+4}}}
{{{0=-4+4}}}
{{{0=0}}}Our answer works and is in the interval 0<(theta)<2pi