If $ x \in \left(0,\frac{\pi}{2}\right)$. Then value of $x$ in $ \frac{3}{\sqrt{2}}\sec x-\sqrt{2}\csc x = 1$ 
If $\displaystyle x \in \left(0,\frac{\pi}{2}\right)$ then find a value of $x$ in $\displaystyle \frac{3}{\sqrt{2}}\sec x-\sqrt{2}\csc x = 1$

$\bf{Attempt:}$ From $$\frac{3}{\sqrt{2}\cos x}-\frac{\sqrt{2}}{\sin x} = 1.$$
$$3\sin x-2\cos x = \sqrt{2}\sin x\cos x$$
$$(3\sin x-2\cos x)^2 = 2\sin^2 x\cos^2 x$$
$$9\sin^2 x+4\cos^2 x-12 \sin x\cos x = 2\sin^2 x\cos^2 x$$
Could some help me to solve it, thanks

Use $\sin{x}=\frac{2t}{1+t^2}$ and $\cos{x}=\frac{1-t^2}{1+t^2}$, where $t=\tan\frac{x}{2}$.
One of roots is $\sqrt2-1$ and the second is very ugly.
For $\tan\frac{x}{2}=\sqrt2-1$ we obtain $\frac{x}{2}=22.5^{\circ}+180^{\circ}k,$ where $k\in\mathbb Z$, 
which gives $x=\frac{\pi}{4}$.
The second root does not give solutions.
