# Trigonometric equations

1. Mar 11, 2013

### phospho

I have shown the first part that they ask for.

For the second part:

let tanθ = t

$tan(3\theta) = \displaystyle\dfrac{tan(2\theta) + tan\theta}{1-tan(\theta)tan(2\theta)} = \dfrac{\frac{2t}{1-t^2} + t}{1 - t(\frac{2t}{1-t^2})}$

hence $t = 2 + \dfrac{3t - t^3}{1-3t^2}$
$t^3 - 3t^2 + t + 1 = (t-1)(t^2 -2t - 1) = 0$

hence $t = 1$, $t = 1 \pm \sqrt{2}$

now I've found the solutions for t = 1, getting θ = pi/4, 5pi/4, but how do I find the solutions for t = 1 ± √2 without using a calculator?

2. Mar 11, 2013

### SammyS

Staff Emeritus
If $\displaystyle\ \tan(\theta)=1\pm\sqrt{2}\,,\$ then what is tan(2θ) ?