Solve the equation on the interval \( 0 \leq 0<2 \pi \) \[ (\tan 0-1)(\cos \theta+1)=0 \] Select the correct choice below and rill in any answer boxes in your choice. A. The solution set is (Simplify your answer. Type an oxact answer, using \( \pi \) as needed. Type your answer in radians. Use integers or B. There is no solution on this interval.

Solve the equation \( (\tan 0-1)(\cos \theta+1)=0 \). Solve the equation by following steps: - step0: Solve for \(\theta\): \(\left(\tan\left(0\right)-1\right)\left(\cos\left(\theta \right)+1\right)=0\) - step1: Simplify: \(-\cos\left(\theta \right)-1=0\) - step2: Move the constant to the right side: \(-\cos\left(\theta \right)=0+1\) - step3: Remove 0: \(-\cos\left(\theta \right)=1\) - step4: Multiply both sides: \(-\cos\left(\theta \right)\left(-1\right)=1\times \left(-1\right)\) - step5: Calculate: \(\cos\left(\theta \right)=1\times \left(-1\right)\) - step6: Calculate: \(\cos\left(\theta \right)=-1\) - step7: Use the inverse trigonometric function: \(\theta =\arccos\left(-1\right)\) - step8: Calculate: \(\theta =\pi \) - step9: Add the period: \(\theta =\pi +2k\pi ,k \in \mathbb{Z}\) The solution to the equation \( (\tan 0-1)(\cos \theta+1)=0 \) on the interval \( 0 \leq \theta < 2\pi \) is \( \theta = \pi + 2k\pi \), where \( k \) is an integer. Therefore, the correct choice is A. The solution set is \( \theta = \pi + 2k\pi \), where \( k \) is an integer.