2tan∅+2=0 solve each equation for 0≤∅≤2π

First, solve for tanØ. 2tanØ+2=0, subtract 2, 2tanØ=-2, divide 2, tanØ=-1. Now, tanØ is only one at 45 degree angles, which in radians is π/4, 3π/4, 5π/4, and 7π/4. For tanØ to be negative, it must be in the second or fourth quadrant. The only answer choices in those quadrants that fit the domain restriction are 3π/4 and 7π/4, making those your answers.