To solve the equation 2sinXcosX + √2cosX - √2sinX - 1 = 0, we can group the terms involving sinX and cosX separately.

We have 2sinXcosX - √2sinX = √2cosX + 1.

Factoring out sinX on the left side and cosX on the right side, we get sinX(2cosX - √2) = √2(cosX + 1).

Dividing both sides by (2cosX - √2), we obtain sinX = √2/(√2 + 2).

This simplifies to sinX = 1/2.

So, X = π/6 or X = 5π/6 (since the sine function repeats every 2π).

Therefore, the solutions for the equation 2sinXcosX + √2cosX - √2sinX - 1 = 0 are X = π/6, 5π/6.