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 November 19th, 2015, 12:47 PM #1 Senior Member     Joined: Dec 2014 From: Canada Posts: 110 Thanks: 4 General Solution I seem to be getting this wrong. Obtain the general solution of the equation $\displaystyle Sin(2x)=Sin(x)$ $\displaystyle 2CosxSinx-Sinx=0$ $\displaystyle Sinx(2cosx-1)=0$ $\displaystyle Sinx=0$ PV = 0 General Solution: $\displaystyle x=pi*k + 0$ $\displaystyle Cosx=\frac{1}{2}$ PV = 60 Degrees General Solution: $\displaystyle x = 360*k +- 60$ k being any integer
 November 19th, 2015, 01:16 PM #2 Math Team   Joined: Jul 2011 From: Texas Posts: 2,837 Thanks: 1479 in the interval $0 \le x < 360^\circ$ , $\cos{x} = \dfrac{1}{2}$ at $x = 60^\circ$ and $x = 300^\circ$ general ... x = 60 + 360k x = 300 + 360k why the switch from radians to degrees for the two zero equations? Thanks from topsquark

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