# Find the x-coordinates of all points on the curve f(x) = sin2x − 2sinx at which the tangent line is horizontal. Enter your answers as a comma-separated list. Use n to represent any integer. Given the fucntions:

f(x)= sin2x - 2sinx

We need to find the points where the tangent is horizontal.

Then we need to find the derivative's zero.

==> f'(x)= 2cos(2x) - 2cos(x) = 0

==> 2cos(2x)= 2cos(x)

==> cos(2x)= cos(x)

==> 2cos^2 x -1 = cosx

==> 2cos^2 c - cos...

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Given the fucntions:

f(x)= sin2x - 2sinx

We need to find the points where the tangent is horizontal.

Then we need to find the derivative's zero.

==> f'(x)= 2cos(2x) - 2cos(x) = 0

==> 2cos(2x)= 2cos(x)

==> cos(2x)= cos(x)

==> 2cos^2 x -1 = cosx

==> 2cos^2 c - cos x -1 = 0

==> (2cosx+1)(cosx -1)  = 0

==> cosx = 1 ==> x = 2npi (n= 0, 1, 2, ...)

==> cosx= -1/2 ==> x = 2pi/3+ 2npi , 4pi/3 + 2npi (n= 0, 1, 2,...)

==> x = { 2npi, 2pi/3+ 2npi, 4pi/3+2npi} ( n= 0, 1, 2, ...}

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