sine 2theta = sine theta
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sin(2t) = sin(t), where 0 < t < 360 degrees
Using the result: sin(2t) = 2sin(t)cos(t), we have
2sin(t)cos(t) = sin(t)
=> 2sin(t)cos(t) - sin(t) = 0
=> sin(t)(2*cos(t) - 1) = 0
=> sin(t) = 0 or cos(t) = 1/2
In the closed interval [0, 2*pi],
sin(t) = 0 for t = 0, pi and 2*pi
cos(t) = 1/2 for t = pi/3 and 5*pi/3
For the open interval (0, 2*pi), the solutions are t = pi, pi/3 or 5*pi/3.
In degrees, that corresponds to:
t = 180 degrees, 60 degrees or 300 degrees
Using the result: sin(2t) = 2sin(t)cos(t), we have
2sin(t)cos(t) = sin(t)
=> 2sin(t)cos(t) - sin(t) = 0
=> sin(t)(2*cos(t) - 1) = 0
=> sin(t) = 0 or cos(t) = 1/2
In the closed interval [0, 2*pi],
sin(t) = 0 for t = 0, pi and 2*pi
cos(t) = 1/2 for t = pi/3 and 5*pi/3
For the open interval (0, 2*pi), the solutions are t = pi, pi/3 or 5*pi/3.
In degrees, that corresponds to:
t = 180 degrees, 60 degrees or 300 degrees