Solving $\frac{(\sec x+\tan x)^2-(\sec 2x+\tan x)^2}{\sin 2x-\sin x}=2$ How can we solve for the value of $x$ in the range $[-\pi,\pi]$?
$$\frac{(\sec x+\tan x)^2-(\sec 2x+\tan x)^2}{\sin 2x-\sin x}=2$$
All i could do is write everything in terms of the elementary function $\sin$ and $\cos$ which gave me the equation
$$\cos^2 2x-\cos^2 x+2\sin x\cos^2 2x-\sin 2x\cos 2x=2\cos^2 x\cos^2 2x\sin x(2\cos x-1)$$
which doesn't seem any good to me. Please enlighten me with a clean solution to this problem.

If there is no typo, it is more than ugly.
Assuming that the denominator cannot be zero, that is to say $x\neq 0$, $x\neq \pm\frac \pi3$ and $x\neq \pm\pi$, using the tangent half-angle substitution $x=2\tan^{-1}(t)$, the lhs write
$$-\frac{ t \left(t^2-3\right) \left(t^2+1\right)^3 \left(2 t^5-3 t^4-12 t^3-2 t^2+2
   t+1\right) } {(t-1)^2 (t+1)^2 \left(t^2-2 t-1\right)^2 \left(t^2+2 t-1\right)^2 \left(3 t^2-1\right) }$$ and we face the "nice" equation
$$5 t^{14}-3 t^{13}-55 t^{12}-2 t^{11}+193 t^{10}+19 t^9-307 t^8+36 t^7+367 t^6+19
   t^5-85 t^4-2 t^3+11 t^2-3 t-1=0$$ which does not factor and does not show rational roots.
Cheating a little, there are only four real roots corresponding to
$$t_1=-0.211814 \quad \quad t_2=0.487927\quad\quad t_3=1.81774\quad\quad t_4=3.04026$$
