Solve $\sin (x) \cos(x) - \cos^{2}(x) = -2$ for complex roots

Question

I was surfing through this website and saw a question asking to solve the equation for its real roots, and wondered whether the complex roots could be found.

$\sin (x) \cos(x) - \cos^{2}(x) = -2$
Solve for complex x.

Here's my attempt:
1. Use the double-angle formula to get both terms on the left side to sin2x and cos2x.
$\sin 2x - \cos 2x = -3 $
2. Use the Harmonic Addition Theorem/R-Formula to obtain
$2^{0.5} \sin (2x-π/4)=-3$
$\sin (2x-π/4)=-3/(2)^{0.5}$
Since it is less than -1, there are no real roots of x.

Let $a=2x-π/4$

Here's WolframAlpha's answer. My attempt and the answer are worlds apart.
How do I fix this thing?

Answer 1

Well, the thing we are trying to solve is the following:

$$\sin\left(x\right)\cos\left(x\right)-\cos^2\left(x\right)=\text{n}\tag1$$

Subtract $\text{n}$ from both sides, and rewrite the LHS of equation $(1)$ a bit different in order to get:

$$-\frac{1}{2}\cdot\left\{1+2\text{n}-\sqrt{2}\cdot\sin\left(\frac{\pi}{4}-2x\right)\right\}=0\tag2$$

Solving for the sine term, gives:

$$\sin\left(\frac{\pi}{4}-2x\right)=-\left(\frac{1}{\sqrt{2}}+\text{n}\sqrt{2}\right)\tag3$$

Take the inverse sine of both sides (then we get two cases):

1. Let $\text{k}_1\in\mathbb{Z}$: $$\frac{\pi}{4}-2x=2\pi\text{k}_1+\pi-\arcsin\left(-\left(\frac{1}{\sqrt{2}}+\text{n}\sqrt{2}\right)\right)\tag4$$
2. Let $\text{k}_2\in\mathbb{Z}$: $$\frac{\pi}{4}-2x=2\pi\text{k}_2+\arcsin\left(-\left(\frac{1}{\sqrt{2}}+\text{n}\sqrt{2}\right)\right)\tag5$$

Using the fact that:

$$\arcsin\left(-x\right)=-\arcsin\left(x\right)\tag6$$

We get:

1. Let $\text{k}_1\in\mathbb{Z}$: $$-2x=2\pi\text{k}_1+\pi-\frac{\pi}{4}+\arcsin\left(\frac{1}{\sqrt{2}}+\text{n}\sqrt{2}\right)\tag6$$
2. Let $\text{k}_2\in\mathbb{Z}$: $$-2x=2\pi\text{k}_2-\frac{\pi}{4}-\arcsin\left(\frac{1}{\sqrt{2}}+\text{n}\sqrt{2}\right)\tag7$$

Last step, divide both sides by $-2$.