The number of solutions of the equation (2x) + (2 - 2 ) (x) + 2 - 1...

Using the identity $\cos(2x) = 1 - 2\sin^2(x)$, the given equation becomes:

$1 - 2\sin^2(x) + (2 - \sqrt{2})\sin(x) + \sqrt{2} - 1 = 0$

$-2\sin^2(x) + (2 - \sqrt{2})\sin(x) + \sqrt{2} = 0$

$2\sin^2(x) - (2 - \sqrt{2})\sin(x) - \sqrt{2} = 0$

Factorizing the quadratic equation:

$2\sin^2(x) - 2\sin(x) + \sqrt{2}\sin(x) - \sqrt{2} = 0$

$2\sin(x)(\sin(x) - 1) + \sqrt{2}(\sin(x) - 1) = 0$

$(2\sin(x) + \sqrt{2})(\sin(x) - 1) = 0$

This gives $\sin(x) = 1$ or $\sin(x) = -\dfrac{1}{\sqrt{2}}$.

Now, we find the number of solutions in the interval $[-\pi, 3\pi]$.

For $\sin(x) = 1$:
In $[-\pi, 3\pi]$, the solutions are $x = \dfrac{\pi}{2}$ and $x = \dfrac{5\pi}{2}$. (2 solutions)

For $\sin(x) = -\dfrac{1}{\sqrt{2}}$:
In $[-\pi, 0]$, the solutions are $x = -\dfrac{\pi}{4}$ and $x = -\dfrac{3\pi}{4}$. (2 solutions)
In $[0, 2\pi]$, the solutions are $x = \dfrac{5\pi}{4}$ and $x = \dfrac{7\pi}{4}$. (2 solutions)
In $[2\pi, 3\pi]$, there are no solutions since $\sin(x) \ge 0$ in this interval.

Total number of solutions = $2 + 2 + 2 = 6$.

Answer: 6