Check the solution to "Solve y3−3y−2=0 using trigonometry."

ashes86047xhz

Answered question

2022-03-20

Solve

y^{3} - 3 y - \sqrt{2} = 0

using trigonometry.

anghoelv1lw

Beginner2022-03-21Added 19 answers

Recall the triplication formula:

\cos 3 α = 4 \cos^{3} α - 3 \cos α

so, if your equation is

y^{3} - 3 y = \sqrt{2}

, you can first set

y = a t

, so

a^{3} t^{3} - 3 a t = \sqrt{2}

and you'd like that

\frac{a^{3}}{3} a = \frac{4}{3}

, so you can take

a = 2 : 8 t^{3} - 6 t = \sqrt{2}

; setting

t = \cos α

, we get

\cos 3 α = \frac{\sqrt{2}}{2}

3 α = \frac{π}{4} + 2 k π

where

k = 0, k = 1

k = 2

. Hence

\frac{π}{12}, \frac{3 π}{4}, \frac{17 π}{12}

are the solutions for

α

.
Now,

2 \cos \frac{π}{12} = 2 \sqrt{\frac{1 + \cos (\frac{π}{6})}{2}} = \sqrt{2 + \sqrt{3}} = \frac{\sqrt{6} + \sqrt{2}}{2}

then

2 \cos \frac{3 π}{4} = - \sqrt{2}

and

2 \cos \frac{17 π}{12} = - 2 \cos \frac{5 π}{12} = - 2 \sqrt{\frac{1 + \cos (5 \frac{π}{6})}{2}}

= - \sqrt{2 - \sqrt{3}} = \frac{\sqrt{6} - \sqrt{2}}{2}

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