Which interval is correct here$?$ The equation $$2\textrm{sin}^2\theta x^2-3\textrm{sin}\theta x+1=0$$ where $\theta \in \left(\frac{\pi}{4},\frac{\pi}{2}\right)$ has one root lying in the interval

$(0,1)$


$(1,2)$


$(2,3)$


$(-1,0)$

I know that if $f(a)$ and $f(b)$ are of opposite signs then at least $1$ or in general odd number of roots of the equation $f(x)=0$ lie between $a$ and $b$. But I am not able to use this piece of information here, maybe because of two variables. I also tried to assume $\textrm{sin}\theta x$ as $y$ but nothing good came out.
Any help is greatly appreciated.
EDIT Answer given is $(1,2)$
 A: Use the quadratic formula to solve for $x$:
$$x_\pm = \frac{3\sin\theta\pm\sqrt{9\sin^2\theta-8\sin^2\theta}}{4\sin^2\theta} = \begin{cases}\frac{1}{\sin\theta} & \text{ or } \\ \frac{1}{2\sin\theta} & \end{cases}$$
because $\sin\theta\in\left( \frac{1}{\sqrt 2},1\right)$ for $\theta\in\left( \frac{\pi}{4},\frac{\pi}{2}\right)$. Thus, the two roots of $x$ lie in the intervals
$$x_+ = \frac{1}{\sin\theta}\in \left( 1,\sqrt 2\right) \subseteq (1,2) \\
x_- = \frac{1}{2\sin\theta}\in \left( \frac{1}{2},\frac{1}{\sqrt 2}\right) \subseteq (0,1)$$
The claim that $x$ only has one root is wrong.
A: $\begin{align}
2\sin^2 \theta x^2 − 3 \sin \theta x +1 & = 0 \\
x & = \frac{3 \sin \theta \pm \sqrt{9 \sin^2 \theta - 8 \sin^2 \theta}}{4 \sin^2 \theta} \\
x & = \frac{3 \pm 1}{4 \sin \theta} \\
x & = \frac{1}{\sin\theta} \; \; \text{or} \; \; x = \frac{1}{2\sin\theta}
\end{align}$
$$ \therefore \, \theta  = (2n+1)\pi - \sin^{-1} \frac{1}{x}, \; 2n\pi + \sin^{-1} \frac{1}{x}, \; \theta = 2n\pi + \sin^{-1} \frac{1}{2x}, \; \theta = (2n+1)\pi - \sin^{-1} \frac{1}{2x} $$
Now, you can finish it.
