2 sin(5θ) cos(5θ) - 2 cos²(5θ) + 3 = 0

The given equation is $2 \sin(5\theta) \cos(5\theta) - 2 \cos^2(5\theta) + 3 = 0$.
We use the identities:
$\sin(2x) = 2 \sin(x) \cos(x)$
$\cos(2x) = 2 \cos^2(x) - 1 \implies 2 \cos^2(x) = \cos(2x) + 1$

Substitute these into the equation, with $x = 5\theta$:

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 $\sin(10\theta) - \cos(10\theta)$.
$R = \sqrt{1^2 + (-1)^2} = \sqrt{1+1} = \sqrt{2}$.
We want to write $\sin(10\theta) - \cos(10\theta)$ as $\sqrt{2} \sin(10\theta - \alpha)$.
Comparing $\sin(10\theta) - \cos(10\theta)$ with $\sqrt{2}(\sin(10\theta)\cos(\alpha) - \cos(10\theta)\sin(\alpha))$, we find $\cos(\alpha) = \frac{1}{\sqrt{2}}$ and $\sin(\alpha) = \frac{1}{\sqrt{2}}$.
This implies $\alpha = \frac{\pi}{4}$.

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 $\boxed{\text{No real solutions}}$.
2 zimekamilika. Zimebaki 3 leo — tuma inayofuata.