# Solve $\tan^2(x)+\tan(x)=2$ for $0\leq x\leq 2\pi$

I am trying to solve the trigonometric equation $$\tan^2(x)+\tan(x)=2$$ for $$0\leq x\leq 2\pi$$. At first glance, I try to rearrange the trigonometric equation into something more manageable such that \begin{align} \frac{\sin^2(x)}{\cos^2(x)}+\frac{\sin(x)}{\cos(x)}&=2 \\ \sin^2(x)+\sin(x)\cos(x)=&2\cos^2(x). \end{align} I do not see how to progress from here. I have also tried using the Pythagorean identity $$\tan^2(x)=\sec^2(x)-1$$, but this did not seem to help. Any suggestions are appreciated.

• $y=\tan x$ satisfies $y^2+y=2$. Commented May 12, 2022 at 12:40

Unfortunately, the algebraic manipulations you have tried so far do not help much. When solving trigonometric equations, it helps to be on the lookout for 'disguised quadratics'. The equation $$\tan^2x+\tan x=2$$ is actually a quadratic equation in $$\tan x$$. This might be clearer if we set $$y=\tan x$$: then we have $$y^2+y=2$$, which is equivalent to $$y^2+y-2=0$$. From here, you can solve this quadratic equation as normal, which gives you the possible values of $$y$$. Then, by substituting $$\tan x$$ back in for $$y$$, you can find the possible values of $$x$$ in the range $$0\le x\le 2\pi$$.
Let $$u = \tan(x)$$
$$\tan^2(x) + \tan(x) = 2 \iff u^2 + u - 2 = 0 \iff (u-1)(u + 2) = 0$$ $$\therefore \tan(x) = -2 \; \text{or} \; \tan(x) = 1$$ $$\therefore x = \pi n - \tan^{-1}(2)\; \text{or} \; x = \frac{\pi}{4} + \pi n, \; n \in \mathbb{Z}$$
In the given domain $$0 \le x \le 2\pi$$, we have $$x = \boxed{\pi - \tan^{-1}(2)} \approx 2.034$$ $$x = \boxed{2\pi - \tan^{-1}(2)} \approx 5.176$$ $$x = \boxed{\frac{\pi}{4}}$$ $$x = \boxed{\frac{5\pi}{4}}$$ $$\blacksquare$$