Solve algebraically for the first four positive values of x. 1 + tan2(x+0.6) = 0

Question

Solve algebraically for the first four positive values of x.
1 + tan2(x+0.6) = 0

I am having issues with one of the algebraic steps, I think. I knowthe answers should be x = 0.275, 0.775, 1.275, and 1.775, but Ihave a negative number as an answer.
I would like a step by step on solving this equationalgebraically.

Here is what I did:
1 + tan2(x+0.6) = 0
tan2(x+0.6) = -1
2(x+0.6) = arctan (-1)
2(x+0.6) = tan-1(-1) + n
x + 0.6 = [tan-1(-1) + n] / 2
x = -0.6 + [tan-1(-1) + n] / 2
x = -0.6 + -0.3926 + n

Explanation / Answer

1+tan(2(x+0.6))=0 Solving that: 1+tan(2(x+0.6))=0 tan(2(x+0.6))=-1 2(x+0.6) = tan-1(-1) 2x + 2(0.6) = -/4 + n 2x = -/4 - 2(0.6) + n 2x = (-0.25 - 1.2 + n) x = (/2) (n - 1.45) x = (n - 1.45)/2 So when you plug in values for n=0,1,2,3... n = 0 x = -0.725 n = 1 x = -0.225 n = 2 x = 0.275 n = 3 x = 0.775 n = 4 x = 1.275 n = 5 x = 1.775 Therefore, your first 4 positive values of x are: x = 0.275, 0.775, 1.275, and 1.775

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