Solve Multiple-Angle Equations...1

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Section 5.3

Note: I had a real hard time solving this multiple-angle problem. It is not clear to me at all. If wrong, can you solve it in full providing steps as you along?

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40

2sin(2x)+sqrt(3)=0
sin(2x)=-sqrt(3)/2

general solutions:

2x= 4pi/3+2pi*k =>x= 4pi/6+2pi/2*k=> x=2pi/3+pi*k
2x=5pi/3+2pi*k=>x=5pi/6+pi*k


Degrees:
x=120°+180°*k
x=150°+180°*k
 
40

2sin(2x)+sqrt(3)=0
sin(2x)=-sqrt(3)/2

general solutions:

2x= 4pi/3+2pi*k =>x= 4pi/6+2pi/2*k=> x=2pi/3+pi*k
2x=5pi/3+2pi*k=>x=5pi/6+pi*k


Degrees:
x=120°+180°*k
x=150°+180°*k

I almost got it right. Why do we add pi•k and not 2pi•k? Is not sine periodic with 2pi?
 

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