# Thread: Solve exactly for tangent?

1. ## Solve exactly for tangent?

Solve the equation exactly for 0 ≤ t ≤ 2pi. tan(2θ) - tanθ = 0.

I know tan 2θ = (2tanθ)/(1-tan^2) and tanθ = sinθ/cosθ

But that is where I get stumped. What would I need to do next? turn the 2(tan θ) on the left to 2(sinθ/cosθ)? Also the tanθ to sinθ/cosθ?

I hope you can help me with this problem and explain to me the process as well. Thank you so much.

2. Originally Posted by nox16
Solve the equation exactly for 0 ≤ t ≤ 2pi. tan(2θ) - tanθ = 0.

I know tan 2θ = (2tanθ)/(1-tan^2) and tanθ = sinθ/cosθ

But that is where I get stumped. What would I need to do next? turn the 2(tan θ) on the left to 2(sinθ/cosθ)? Also the tanθ to sinθ/cosθ?

I hope you can help me with this problem and explain to me the process as well. Thank you so much.
You don't need to change the tan into sines and cosines.

$\tan (2\theta) - \tan \theta = \frac{2\tan\theta}{1-tan^2 \theta} - \tan \theta = 0$

Factorising $\tan \theta$
$\tan \theta \left( \frac{2}{1-tan^2\theta} - 1 \right) = 0$

Now $\tan \theta = 0$ or $\left( \frac{2}{1-tan^2\theta} - 1 \right) = 0$

However
$\left( \frac{2}{1-tan^2\theta} - 1 \right) = 0$ simplifies to $tan^2\theta = -1$. But a square of a real number cannot be negative, so only $\tan \theta = 0$

$\tan \theta = 0$ occurs three times in the interval specified by this question. Can you find them?

3. Yes I can thank you so much for helping me on this. I was going crazy trying to convert that thing into sin or cos so I could solve it. I never thought about leaving it just in tangent form and then solving it from there. Excellent work on showing your work, it was really helpful.