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Yes, it is possible for a range of values of x.
To determine these values, we'll have to solve the above equation.
First, to emphasize the fact that we are talking about an equation, we'll move the terms sin 2x, to the left side.
sin x - sin 2x = 0
Now, it is obvious that we'll have to determine the values of x that cancel the equation.
First of all, we'll try to calculate sin 2x = sin (x+x)
sin (x+x) = sin x*cos x + cos x*sin x
sin (x+x) = 2sin x*cos x
2sin x*cos x - sin x = 0
sin x(2cos x - 1)=0
If a product of 2 factors is zero, that means that one of the 2 factors is zero.
sin x = 0
This is an elementary equation.
x= (-1)^k*arcsin 0 + k*pi
We'll put the second factor as being zero: 2cos x - 1=0
2cos x = 1
cos x = 1/2
x = (-1)^k*arcsin (1/2)+k*pi
So, the identity sin x = sin 2x is true for x = (-1)^k*pi/6 + k*pi.
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