Solve {l}{tan(3phi-4)=-1/2}{a={(2)}}{text{Solvefor}btext{where}}{b=a}

$$\left. \begin{array} { l } { \tan(3 \phi - 4) = -\frac{1}{2} }\\ { a = {(2)} }\\ { \text{Solve for } b \text{ where} } \\ { b = a } \end{array} \right.$$

Solve for ϕ, a, b (complex solution)

$$ϕ=\frac{2\pi n_{1}+4-\arctan(\frac{1}{2})}{3}\text{, }n_{1}\in \mathrm{Z}\text{, }a=2\text{, }b=2$$

$$ϕ=\frac{2\pi n_{2}+4-\pi -\arctan(\frac{1}{2})}{3}\text{, }n_{2}\in \mathrm{Z}\text{, }a=2\text{, }b=2$$

Solve for ϕ, a, b

$$ϕ=\frac{2\pi n_{1}-\arcsin(\frac{\sqrt{5}}{5})+4-\pi }{3}\text{, }n_{1}\in \mathrm{Z}\text{, }a=2\text{, }b=2$$

$$ϕ=\frac{2\pi n_{2}-\arcsin(\frac{\sqrt{5}}{5})+4}{3}\text{, }n_{2}\in \mathrm{Z}\text{, }a=2\text{, }b=2$$