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November 19th, 2015, 12:47 PM   #1
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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
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November 19th, 2015, 01:16 PM   #2
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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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