What is the general solution for tan(θ)=(m^2-m^1)/(1+m^2*m^1),m^2 ?

The general solution for tan(θ)=(m^2-m^1)/(1+m^2*m^1),m^2 is

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tan(θ)=(m^2-m^1)/(1+m^2*m^1),m^2

tan (θ) = \frac{m ^{2} - m ^{1}}{1 + m ^{2} \cdot m ^{1}}, m^{2}

θ = arctan (\frac{m ^{2} - m}{m ^{3} + 1}), θ = arctan (\frac{m ^{2} - m}{1 + m ^{3}}) + π, θ = arctan (\frac{m ^{2} - m}{1 + m ^{3}}) + 2 π, θ = arctan (\frac{m ^{2} - m}{1 + m ^{3}}) + 3 π, θ = arctan (\frac{m ^{2} - m}{1 + m ^{3}}) + 4 π, θ = arctan (\frac{m ^{2} - m}{1 + m ^{3}}) - π, θ = arctan (\frac{m ^{2} - m}{1 + m ^{3}}) - 2 π, θ = arctan (\frac{m ^{2} - m}{1 + m ^{3}}) - 3 π, θ = arctan (\frac{m ^{2} - m}{1 + m ^{3}}) - 4 π

Solution steps

tan (θ) = \frac{m ^{2} - m ^{1}}{1 + m ^{2} m ^{1}}, m^{2}

Simplify

\frac{m ^{2} - m ^{1}}{1 + m ^{2} m ^{1}} : \frac{m ^{2} - m}{1 + m ^{3}}

Apply trig inverse properties

θ = arctan (\frac{m ^{2} - m}{1 + m ^{3}}) + πn

Solutions for the range

m^{2}

θ = arctan (\frac{m ^{2} - m}{m ^{3} + 1}), θ = arctan (\frac{m ^{2} - m}{1 + m ^{3}}) + π, θ = arctan (\frac{m ^{2} - m}{1 + m ^{3}}) + 2 π, θ = arctan (\frac{m ^{2} - m}{1 + m ^{3}}) + 3 π, θ = arctan (\frac{m ^{2} - m}{1 + m ^{3}}) + 4 π, θ = arctan (\frac{m ^{2} - m}{1 + m ^{3}}) - π, θ = arctan (\frac{m ^{2} - m}{1 + m ^{3}}) - 2 π, θ = arctan (\frac{m ^{2} - m}{1 + m ^{3}}) - 3 π, θ = arctan (\frac{m ^{2} - m}{1 + m ^{3}}) - 4 π

0 = - 1 - tan^{2} (θ) + 2 tan (θ)

u, x = tanh (u)

3 + 2 cos (x) = 0

2 cos^{2} (θ) - 7 cos (θ) = - 3

tan^{2} (x) = 3 - 2 tan (x)

What is the general solution for tan(θ)=(m^2-m^1)/(1+m^2*m^1),m^2 ?
The general solution for tan(θ)=(m^2-m^1)/(1+m^2*m^1),m^2 is