Trigonometric equations

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I have shown the first part that they ask for.

For the second part:

let tanθ = t

[itex] 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})} [/itex]

hence [itex] t = 2 + \dfrac{3t - t^3}{1-3t^2} [/itex]
[itex] t^3 - 3t^2 + t + 1 = (t-1)(t^2 -2t - 1) = 0 [/itex]

hence [itex] t = 1 [/itex], [itex] t = 1 \pm \sqrt{2} [/itex]

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?
 

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SammyS
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SVxIMJY.png


I have shown the first part that they ask for.

For the second part:

let tanθ = t

[itex] 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})} [/itex]

hence [itex] t = 2 + \dfrac{3t - t^3}{1-3t^2} [/itex]
[itex] t^3 - 3t^2 + t + 1 = (t-1)(t^2 -2t - 1) = 0 [/itex]

hence [itex] t = 1 [/itex], [itex] t = 1 \pm \sqrt{2} [/itex]

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 [itex]\displaystyle\ \tan(\theta)=1\pm\sqrt{2}\,,\ [/itex] then what is tan(2θ) ?
 

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