Find the general solution for this equation: \[ -\tan \left(2 x-45^{\circ}\right)+2=5 \] INSTRUCTION: Round your answers to two decimal places, and if there is more than o the answers with the \( \because{ }^{\circ} \) symbol. Answer: \( x=\square ; k \in \mathbb{Z} \)

To solve the equation \(-\tan(2x - 45^\circ) + 2 = 5\), we first isolate the tangent function: \[ -\tan(2x - 45^\circ) = 3 \] \[ \tan(2x - 45^\circ) = -3 \] Now, we can find the general solution for \(\tan(\theta) = -3\): \[ 2x - 45^\circ = \tan^{-1}(-3) + k \cdot 180^\circ \] Calculating \(\tan^{-1}(-3)\): Using a calculator, \(\tan^{-1}(-3) \approx -71.57^\circ\). Thus, we have: \[ 2x - 45^\circ = -71.57^\circ + k \cdot 180^\circ \] Now, solve for \(x\): \[ 2x = -71.57^\circ + 45^\circ + k \cdot 180^\circ \] \[ 2x = -26.57^\circ + k \cdot 180^\circ \] \[ x = -13.285^\circ + \frac{k \cdot 180^\circ}{2} \] \[ x = -13.29^\circ + 90^\circ k \] In rounding to two decimal places, we get: \[ x \approx -13.29^\circ + 90^\circ k \] Therefore, the general solution is: \[ x = -13.29^\circ + 90^\circ k; \quad k \in \mathbb{Z} \]