Given z = (5)/(2-i) - (3+2i), find z2 and its modulus and argument. Then, solve the equation (z)/(1+2i) + (z-1)/(5) = (1)/(1-2i) for z.

$$z = (2+i) - (3+2i) = 2+i-3-2i = (2-3) + (1-2)i = -1-i$$

So, $z$ in the form $x+iy$ is $\boxed{-1-i}$.

For $z^2 = 0+2i$, the modulus is:

$$|z^2| = \sqrt{0^2 + 2^2} = \sqrt{4} = 2$$

So, $|z^2| = \boxed{2}$.

For $z^2 = 2i$, the complex number lies on the positive imaginary axis.

$$\arg z^2 = \frac{\pi}{2}$$

So, $\arg z^2 = \boxed{\frac{\pi}{2}}$.

$$\frac{z}{1+2i} = \frac{z(1-2i)}{(1+2i)(1-2i)} = \frac{z(1-2i)}{1^2 - (2i)^2} = \frac{z(1-2i)}{1+4} = \frac{z(1-2i)}{5}$$ $$\frac{1}{1-2i} = \frac{1(1+2i)}{(1-2i)(1+2i)} = \frac{1+2i}{1^2 - (2i)^2} = \frac{1+2i}{1+4} = \frac{1+2i}{5}$$

$$z(1-2i) + (z-1) = 1+2i$$ $$z - 2iz + z - 1 = 1+2i$$ $$2z - 2iz - 1 = 1+2i$$ $$z(2-2i) = 1+2i+1$$ $$z(2-2i) = 2+2i$$

$$z = \frac{2+2i}{2-2i} = \frac{2(1+i)}{2(1-i)} = \frac{1+i}{1-i}$$

Multiply by the conjugate of the denominator:

$$z = \frac{1+i}{1-i} \times \frac{1+i}{1+i} = \frac{(1+i)^2}{1^2 - i^2} = \frac{1+2i+i^2}{1 - (-1)} = \frac{1+2i-1}{2} = \frac{2i}{2} = i$$

So, $z=i$.

Substitute $z=i$:

$$(p+iq)i = 3+4i$$ $$pi + i^2q = 3+4i$$ $$pi - q = 3+4i$$

Rearrange to group real and imaginary parts:

$$-q + pi = 3+4i$$

Equating the real parts:

$$-q = 3 \implies q = -3$$

Equating the imaginary parts:

$$p = 4$$

So, $p = \boxed{4}$ and $q = \boxed{-3}$.

$$(2\cos^2 x - 1) + 1 = \sin 2x$$ $$2\cos^2 x = \sin 2x$$

$$2\cos^2 x - 2\sin x \cos x = 0$$ $$2\cos x (\cos x - \sin x) = 0$$

Case 1: $2\cos x = 0$

$$\cos x = 0$$

The general solution for $\cos x = 0$ is $x = \frac{\pi}{2} + n\pi$, where $n \in \mathbb{Z}$.

Case 2: $\cos x - \sin x = 0$

$$\cos x = \sin x$$

Divide by $\cos x$ (assuming $\cos x \neq