Assuming you mean that the pi is not within the sin(2pi),
its a vertical shift of +pi
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Shift = +2
a normal sine curve exists with the formula Asin(Bx+C)+D. The formula to derive a phase shift would be such: 2pi/B (for whatever value B exists at). Thus, for a normal sine curve (sin(x) we would get 2pi/1, and arrive at 2pi for the period.
The graph of y = A sin (Bx + C) is obtained by horizontally shifting the graph of y = A sin Bx, where 1) Amplitude = |A|, the maximum of y is A, if A > 0 and -(-A) if A < 0 The minimum of y is -A if A > 0 and A if A < 0 2) Period = (2pi)/B 3)Phase shift = C/B So that the starting point of the cycle is shifted from x = 0 to x = C/B. If C/B > 0 the shift is to the right. If C/B < 0 the shift is to the left.
the period is 2pi. period is 2pi/b and the formula is y=AsinBx.
Sin(2x) + Sin(x) = 0 2Sin(x)Cos(x) + Sin(x) = 0 Factor 'Sin*(x)' Sin(x) (2Cos(x) + 1 = 0 Hence Sin(x) = 0 x = 0 , 180(pi) , 360(2pi) et.seq., & 2Cos(x) = -1 Cos(x) = -1/2 = -0.5 x = 120, 150, 480, et.seq.,