Сonvert trigonometric equation to quadratic equation I am struggling to know how to transform this equation as it involves more than one type of trigonometric function, I know how to do it with one repeated function. 
Question:
$\sin^2 \theta/2$ + $\sin \theta$ + $1$ = $0$
In must be transform in pure quadratic form:
$t^2$ + $t$ + $1$ = $0$

This is what is sometime called a "trick question" since it does not require you to solve a quadratic equation.
Notice that if $\sin^2\left(\dfrac{\theta}{2}\right)+\sin\theta+1=0$, then it follows that $\sin^2\left(\dfrac{\theta}{2}\right)=-\sin\theta-1\ge0$. So it must be the case that $\sin\theta\le-1$. So $\sin\theta=-1$.
Therefore one would think that $\theta=-\dfrac{\pi}{2}+2\pi n$. But this will contain extraneous solutions. For example, $-\dfrac{\pi}{2}$ is not a solution since
$$ \sin^2\left(-\frac{\pi}{4}\right)+\sin\left(-\frac{\pi}{2}\right)+1\ne0 $$
This raises the question "Are there any values of $n$ which yield a solution?"
\begin{eqnarray}
&&\sin^2\left(-\frac{\pi}{4}+\pi n\right)+\sin\left(-\frac{\pi}{2}+2\pi n\right)+1\\
&=&-\frac{\sqrt{2}}{2}\cos(\pi n)+0-\cos(2\pi n)+0\\
&=&-\frac{\sqrt{2}}{2}\cos(\pi n)-1
\end{eqnarray}
But, since $\cos(\pi n)=\pm 1$ there are no values of $n$ which will give $0$.
So the equation has no solutions.
ADDENDUM
If it is the case that OP intended to write $\dfrac{\sin^2\theta}{2}+\sin\theta+1=0$, that also has no solutions since
\begin{eqnarray}
&&\sin^2\theta+2\sin\theta+2\\
&=&(\sin\theta+1)^2+1\ge1
\end{eqnarray}
SECOND ADDENDUM
Here is an even faster way to show that there are no solutions.
Rewrite the equation using the identity $\sin^2\left(\dfrac{\theta}{2}\right)=\dfrac{1-\cos\theta}{2}$ to obtain
$$ 2\sin\theta-\cos\theta+3=0 $$
$$ \sqrt{5}\left(\frac{2}{\sqrt{5}}\sin\theta-\frac{1}{\sqrt{5}}\cos\theta\right)+3=0$$
Letting $\sin\psi=\dfrac{1}{\sqrt{5}}$ this can be re-written
$$ \sqrt{5}\sin(\theta-\psi)+3=0 $$
But $\sqrt{5}\sin(\theta-\psi)+3\ge3-\sqrt{5}>0$. So there are no solutions.
