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

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

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.

To find the values of x between 0 and 2π (inclusive) that satisfy the equation sin(4x) = sin(x), we can use the properties of the sine function and solve it algebraically.

We know that sin(x) = sin(y) if either x = y + 2nπ or x = π - y + 2nπ, where n is an integer.

Applying this property to the equation sin(4x) = sin(x), we have two possible cases:

Case 1: 4x = x + 2nπ
Simplifying the equation, we get:
4x - x = 2nπ
3x = 2nπ
x = (2nπ)/3

Case 2: 4x = π - x + 2nπ
Simplifying the equation, we get:
4x + x = π + 2nπ
5x = π + 2nπ
x = (π + 2nπ)/5

Now, let's substitute the integer values of n from 0 to 8 into both cases to find the values of x between 0 and 2π that satisfy the equation:

For Case 1:
n = 0: x = (2 * 0 * π)/3 = 0
n = 1: x = (2 * 1 * π)/3 = 2π/3
n = 2: x = (2 * 2 * π)/3 = 4π/3

For Case 2:
n = 0: x = (π + 2 * 0 * π)/5 = π/5
n = 1: x = (π + 2 * 1 * π)/5 = 3π/5
n = 2: x = (π + 2 * 2 * π)/5 = 5π/5 = π
n = 3: x = (π + 2 * 3 * π)/5 = 7π/5
n = 4: x = (π + 2 * 4 * π)/5 = 9π/5
n = 5: x = (π + 2 * 5 * π)/5 = 11π/5
n = 6: x = (π + 2 * 6 * π)/5 = 13π/5
n = 7: x = (π + 2 * 7 * π)/5 = 15π/5 = 3π

Therefore, the nine values of x (in radians) between 0 and 2π that satisfy the equation sin(4x) = sin(x) are:
0, 2π/3, 4π/3, π/5, 3π/5, π, 7π/5, 9π/5, and 3π.