Solve tan2x+tan(pi/4-x)=a

$$\tan 2 x + \tan ( \frac { \pi } { 4 } - x ) = a$$

Solve for a (complex solution)

$$\left\{\begin{matrix}a=\frac{2e^{\frac{\pi i-12ix}{4}}\left(i-ie^{\frac{4ix+\pi i}{2}}\right)}{e^{\frac{4ix+\pi i}{4}}+e^{\frac{-12ix+\pi i}{4}}+e^{\frac{-4ix+3i\pi }{4}}+e^{\frac{-20ix+3i\pi }{4}}}\text{, }&-\frac{1}{e^{ix+\frac{\pi i}{4}}}-\frac{1}{e^{-3ix+\frac{\pi i}{4}}}-e^{ix+\frac{\pi i}{4}}-e^{-3ix+\frac{\pi i}{4}}\neq 0\text{ and }\nexists n_{1}\in \mathrm{Z}\text{ : }x=\frac{\pi n_{1}}{2}+\frac{\pi }{4}\\a\in \mathrm{C}\text{, }&\exists n_{2}\in \mathrm{Z}\text{ : }x=\pi n_{2}+\frac{3\pi }{4}\text{ and }-\frac{1}{e^{ix+\frac{\pi i}{4}}}-\frac{1}{e^{-3ix+\frac{\pi i}{4}}}-e^{ix+\frac{\pi i}{4}}-e^{-3ix+\frac{\pi i}{4}}=0\text{ and }\nexists n_{1}\in \mathrm{Z}\text{ : }x=\frac{\pi n_{1}}{2}+\frac{\pi }{4}\end{matrix}\right.$$

Solve for a

$$a=\frac{1}{\cos(2x)}$$

$$\nexists n_{1}\in \mathrm{Z}\text{ : }x=\frac{\pi n_{1}}{2}+\frac{\pi }{4}$$

Solve for x

$$x=\frac{-\arccos(\frac{1}{a})+2\pi n_{1}}{2}\text{, }n_{1}\in \mathrm{Z}$$

$$x=\frac{\arccos(\frac{1}{a})+2\pi n_{2}}{2}\text{, }n_{2}\in \mathrm{Z}\text{, }|a|\geq 1$$