# Solve for θ. tan(2θ)−tan(2θ)tan^2(θ)=2

Solve for $$\displaystyleθ$$.
$$\displaystyle{\tan{{\left({2}θ\right)}}}−{\tan{{\left({2}θ\right)}}}{{\tan}^{{2}}{\left(θ\right)}}={2}$$

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grbavit

Firstly, notice that $$\displaystyle{\tan{{\left({2}\theta\right)}}}$$ appears in both terms on the left so let's factor that out.
$$\displaystyle{\tan{{\left({2}θ\right)}}}{\left[{1}−{{\tan}^{{2}}{\left(θ\right)}}\right]}={2}$$
Now, we need the arguments of our tantan functions to be the same if we have any hopes of simplifying the left-hand side. So we'll use the double angle formula to convert that $$\displaystyle{\tan{{\left({2}θ\right)}}}$$ into an expression in terms of $$\tan(θ)$$. So remember that the double angle formula for tantan is
$$\displaystyle{\tan{{\left({2}θ\right)}}}=\frac{{{2}{\tan{{\left(θ\right)}}}}}{{{1}-{\left({{\tan}^{{2}}{\left(θ\right)}}\right)}}}$$
Using that, we get $$\displaystyle\to{\tan{{\left(θ\right)}}}={1}$$
Here we look at the unit circle. Note that $$\displaystyle{\tan{{\left(θ\right)}}}={1}$$ means the same thing as $$\displaystyle{\sin{{\left(θ\right)}}}={\cos{{\left(θ\right)}}}$$ which means the same thing as the x component = the y component on the unit circle. That only occurs at two angles between 0 and $$\displaystyle{2}\pi$$ radians:$$\displaystyle\frac{\pi}{{4}}\ \text{ radians }\ \ \text{ and }\ {5}\frac{\pi}{{4}}$$ radians.

Remember though that the angle doesn't necessarily have to be between 0 and $$2\pi$$ radians so we need to add $$2\pi k$$ to both of those angles. Hence the full solution is $$\frac{\pi}{4}+\frac{2}{\pi k}$$ or $$\frac{5\pi}{4+2\pi k}$$ for every integer k, or more succinctly

$$\frac{\pi}{4}+\pi k$$ for every integer k