How to Solve for θ When tan(θ) = 1 ± √2?

[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?

 

[SammyS]

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?

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