The tangent(A) is = 1 at 45 degrees, 225 degrees, 405 degrees and 585 degrees
So, let A = 2Θ + 15
And setting 2Θ + 15 equal to the above values, we have
2Θ + 15 = 45 and solving for Θ we have Θ = 15 degrees
2Θ + 15 = 225 Θ = 105 degrees
2Θ + 15 = 405 Θ = 195 degrees
2Θ + 15 = 585 Θ = 285 degrees
And those are all the values that make the equation true for 0 ≤ Θ ≤ 360
You can see the graph of the solutions here.....https://www.desmos.com/calculator/xliaolg9le
The tangent(A) is = 1 at 45 degrees, 225 degrees, 405 degrees and 585 degrees
So, let A = 2Θ + 15
And setting 2Θ + 15 equal to the above values, we have
2Θ + 15 = 45 and solving for Θ we have Θ = 15 degrees
2Θ + 15 = 225 Θ = 105 degrees
2Θ + 15 = 405 Θ = 195 degrees
2Θ + 15 = 585 Θ = 285 degrees
And those are all the values that make the equation true for 0 ≤ Θ ≤ 360
You can see the graph of the solutions here.....https://www.desmos.com/calculator/xliaolg9le