# Maths

Find the nine values for x ( in radians) with 0=/< x </= 2Pi... so that sin4x = sinx ?

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1. sin4x = sinx
2sin2x cos2x = sinx
2(2sinxcosx)(2cos^2x - 1) = sinx
divide by sinx
4cosx(2cos^2x - 1) = 1
8cos^3x - 4cosx - 1 = 0

let cosx = t

8t^3 - 4t - 1 = 0
using a few trials, I got t = -1/2
and by division I got
(2t+1)(4t^2 - 2t - 1) = 0
t = -1/2 or t = (1+√5)/4 or t=(1-√5)/4
cosx = -1/2 or cosx = (1+√5)/4 or cosx=(1-√5)/4
x = 2π/3 or 4π/3 (from cosx=-1/2)
x = .6283 or 5.6549 from cosx = (1+√5)/4
x = 1.8850 or 4.3983 from cosx = (1-√5)/4

in the interval 0 ≤ x < 2π there are only 6 solutions, not 9

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2. Should not have just divided by sinx and dropped it.

2(2sinxcosx)(2cos^2x - 1) = sinx
2(2sinxcosx)(2cos^2x - 1) - sinx = 0
sinx(8cos^3x - 4cosx - 1) = 0

so we also have sinx = 0
x = 0, π or 2π

the other part stays the same.
So in addition to the 6 answers above we have these 3 new ones,
and YES, there are 9 solutions.

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