Tangents

Notice that we can express the given equation in terms of $\tan\left(4\theta\right) = \dfrac{4\tan(\theta) - 4\tan^3(\theta)}{1 - 6\tan^2(\theta) + \tan^4(\theta)}$ since by arranging terms, $\begin{array}{rl} \tan^4(\theta) - 6\tan^2(\theta) + 1 &= 4\tan^3(\theta) - 4\tan(\theta)\\ 1 &= \dfrac{4\tan^3(\theta) - 4\tan(\theta)}{1 - 6\tan^2(\theta) + \tan^4(\theta)}\\ -1 &= \dfrac{4\tan(\theta) - 4\tan^3(\theta)}{1 - 6\tan^2(\theta) + \tan^4(\theta)}\\ \tan\left(-\dfrac{\pi}{4}\right) &= \dfrac{4\tan(\theta) - 4\tan^3(\theta)}{1 - 6\tan^2(\theta) + \tan^4(\theta)}\\ \Longrightarrow 4\theta &= \pi k - \dfrac{3\pi}{4} \quad k \in \mathbb{Z}\\ \theta &= \dfrac{\pi k}{4} - \dfrac{3\pi}{16} \end{array}$ Between $0 \leq \theta \leq 2\pi$ , the set of solutions is $\theta = \left\{\dfrac{\pi}{16}, \dfrac{5\pi}{16}, \dfrac{9\pi}{16}, \dfrac{13\pi}{16}, \dfrac{17\pi}{16}, \dfrac{21\pi}{16}, \dfrac{25\pi}{16}, \dfrac{29\pi}{16}\right\}$ all in which $\tan(\theta) \neq 1 \pm \sqrt{2}$ (see Note ).. Thus, the sum of all solutions is $\dfrac{15\pi}{2}\text{ radians } = \boxed{1350^{\circ}}$

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Note

To show that the solutions do not give $1 \pm \sqrt{2}$ , solve the given equation as if we were to solve a polynomial. Let $x = \tan(\theta)$ . Then, $x^4 + 4x^3 - 6x^2 - 4x + 1 = 0$ Dividing both sides by $x^2$ , $x^2 + 4x - 6 - \dfrac{4}{x} + \dfrac{1}{x^2} = 0$ which can be expressed as $\begin{array}{rl} \left(x^2 + \dfrac{1}{x^2}\right) + 4\left(x - \dfrac{1}{x}\right) - 6 &= 0\\ \left(x^2 + \left(-\dfrac{1}{x}\right)^2 - 2\right) + 4\left(x - \dfrac{1}{x}\right) - 6 + 2 &= 0\\ \left(x - \dfrac{1}{x}\right)^2 + 4\left(x - \dfrac{1}{x}\right) - 4 &= 0 \end{array}$ Let $u = x - \dfrac{1}{x}$ , so $u^2 + 4u - 4 = 0$ where $u = \{-2 \pm \sqrt{2}\}$ Then, $\begin{array}{rl} x - \dfrac{1}{x} &= u\\ x^2 - 1 &= ux\\ x^2 - ux - 1 &= 0\\ x &= \dfrac{u \pm \sqrt{u^2 + 4}}{2}\\ \end{array}$ which thus yields $\begin{array}{rl} \tan(\theta_1) &= -1 + \sqrt{2} + \sqrt{4 - 2\sqrt{2}}\\ \tan(\theta_2) &= -1 + \sqrt{2} - \sqrt{4 - 2\sqrt{2}}\\ \tan(\theta_3) &= -1 - \sqrt{2} - \sqrt{4 + 2\sqrt{2}}\\ \tan(\theta_4) &= -1 - \sqrt{2} + \sqrt{4 + 2\sqrt{2}}\\ \end{array}$