$\sin 2x - \sin^2x = 20\sin x + 10 \cos x - 1$ with $90\leq x \leq 180$ solve $\sin2x - \sin^2x = 20\sin x + 10 \cos x - 1$ with $90\leq x \leq 180$


I arrived at $2\sin x \cos x + \cos^2 x = 20\sin x + 10 \cos x $ and resulting in $(2\sin x + \cos x)(\cos x - 10) = 0$ but i am stuck at finding the x satisfying $(2\sin x + \cos x) = 0$. Maybe i was wrong somewhere. What should i do?
It's enough to solve the following equation:
$$2\sin{x}+\cos{x}=0.$$
Now, if $\cos{x}=0,$ so $\sin{x}=0,$ which gives $\sin^2x+\cos^2x=0,$ which is a contradiction.

Thus, it's $$\tan{x}=-\frac{1}{2}$$ or
$$x=-\arctan\frac{1}{2}+180^{\circ}k,$$ where $k\in\mathbb Z,$ which gives
$$x=180^{\circ}-\arctan\frac{1}{2}.$$
It's enough to solve the following equation:
$$2\sin{x}+\cos{x}=0.$$
Now, if $\cos{x}=0,$ so $\sin{x}=0,$ which gives $\sin^2x+\cos^2x=0,$ which is a contradiction.

Thus, it's $$\tan{x}=-\frac{1}{2}$$ or
$$x=-\arctan\frac{1}{2}+180^{\circ}k,$$ where $k\in\mathbb Z,$ which gives
$$x=180^{\circ}-\arctan\frac{1}{2}.$$