2sin(50)cos(50) - 2cos²(50) + 3 = 0

The given equation is $2 \sin(\theta) \cos(\theta) - 2 \cos^2(\theta) + 3 = 0$.
We use the identities:
$\sin(2\theta) = 2 \sin(\theta) \cos(\theta)$
$\cos(2\theta) = 2 \cos^2(\theta) - 1 \implies 2 \cos^2(\theta) = \cos(2\theta) + 1$

Substitute these into the equation:

For an expression $a \sin(x) + b \cos(x)$, we can write it as $R \sin(x + \alpha)$, where $R = \sqrt{a^2 + b^2}$.
Here, $a=1$ and $b=-1$ for $\sin(2\theta) - \cos(2\theta)$.
$R = \sqrt{1^2 + (-1)^2} = \sqrt{1+1} = \sqrt{2}$.
We want to write $\sin(2\theta) - \cos(2\theta)$ as $\sqrt{2} \sin(2\theta - \alpha)$.
Comparing $\sin(2\theta) - \cos(2\theta)$ with $\sqrt{2}(\sin(2\theta)\cos(\alpha) - \cos(2\theta)\sin(\alpha))$, we have:
$\sqrt{2}\cos(\alpha) = 1 \implies \cos(\alpha) = \frac{1}{\sqrt{2}}$
$\sqrt{2}\sin(\alpha) = 1 \implies \sin(\alpha) = \frac{1}{\sqrt{2}}$
This implies $\alpha = \frac{\pi}{4}$ (or $45^\circ$).

Divide by $\sqrt{2}$:

The range of the sine function is $[-1, 1]$.
Since $-\sqrt{2} \approx -1.414$, which is less than $-1$, there is no real angle whose sine is $-\sqrt{2}$.
Therefore, there are no real solutions for $\theta$.

The final answer is $\text{No real solutions}$.