If $\sec \theta + \tan \theta =x$, then find the value of $\sin \theta$. If $\sec \theta + \tan \theta =x$, then find the value of $\sin \theta$.
$$\sec \theta + \tan \theta = x$$
$$\dfrac {1}{\cos \theta }+\dfrac {\sin \theta }{\cos \theta }=x$$
$$\dfrac {1+\sin \theta }{\sqrt {1-\sin^2 \theta }}=x$$
$$1+\sin \theta =x\sqrt {1-\sin^2 \theta }$$
$$1+2\sin \theta + \sin^2 \theta = x^2-x^2 \sin^2 \theta $$
$$x^2 \sin^2 \theta + \sin^2 \theta + 2\sin \theta = x^2-1$$
$$\sin^2 \theta (x^2+1) + 2\sin \theta =x^2-1$$

From where you are:
You obtained a quadratic function in $\sin(\theta)$. Perform the substitution $u =\sin(\theta)$. 
We obtain the quadratic (in $u$):
$$(x^2+1)u^2 + 2u - x^2 +1 = 0$$ 
$$\Rightarrow u_{1,2} = \frac{- 2 \pm \sqrt{4 - 4(x^2+1)(1-x^2)}}{2(x^2+1)}$$
$$ = \frac{- 2 \pm \sqrt{4 + 4(x^4 -1)}}{2(x^2+1)}$$
$$ = \frac{- 2 \pm \sqrt{4x^4}}{2(x^2+1)}$$
$$ = \frac{- 2 \pm 2x^2}{2(x^2+1)}$$
$$ = \frac{- 1 \pm x^2}{x^2+1}$$
Therefore, 
$$\sin(\theta)_{1,2} = \frac{- 1 \pm x^2}{x^2+1}$$
One of those solutions will not work out.
This happened because you squared the equation multiple times and we know that $a = b$ is not equivalent with $a^2 = b^2$, so you should fill in both solutions in the original expression and see which one works and which one doesn't.
