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 2CosxSinxSinx=0 $ $\displaystyle Sinx(2cosx1)=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: 2,923 Thanks: 1519 
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? 

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