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Find all angles, 0≤A<360, that satisfy the equation below, to the nearest 10th of a degree.

2tanA+6=tanA+3

1 answer

2tanA+6=tanA+3
tanA = -3
so A must be in II or IV
using tanx = +3
x = 71.565°

so A = 180-71.564 = appr 108.4
or
A = 360 - 71.565 = appr 288.4°

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