Nested fractions trig equation from the Czech version of SAT

$\begin{aligned} \frac 23 \cos x & = \frac 1{2\tan x + \frac 1{2\tan x + \frac 1{2\tan + \cdots}}} \\ & = \frac 1{2\tan x + \frac 23 \cos x} \\ \frac 3{2\cos x} & = 2 \tan x + \frac 23 \cos x \\ 9 & = 12 \sin x + 4 \cos^2 x \\ 9 & = 12 \sin x + 4 (1-\sin^2 x) \\ 4\sin^2 x - 12\sin x + 5 & = 0 \\ (2\sin x - 1)(2\sin x - 5) & = 0 & \small \blue{\text{Since }\sin x \le 1} \\ \implies \sin x & = \frac 12 \\ \implies x & = \boxed {150}^\circ & \small \blue{\text{For }x \in (90^\circ, 360^\circ)} \end{aligned}$

I'm confused whether the interval is open or closed. So I assume that it is open.

R. H. S. of the given equation is equal to $\sec x-\tan x$ . So $2\cos^2 x=3-3\sin x$

$\implies 2\sin^2 x-3\sin x+1=0\implies \sin x=1$ or $\sin x=\dfrac {1}{2}$ .

$\implies x=90\degree$ or $x=150\degree$ . Assuming open interval, $x=\boxed {150\degree}$ .