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    I can literally do all of them but I can't seem to understand this one??

    http://prntscr.com/bgeogr

    Thanks
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    Not 100% sure, but arcsin(0) = 0

    0 + 2(pi)n = x + pi/6

    x = 2pi(n) - pi/6

    The n is because sin is periodic every 2pi radians, this is my interpretation of the answer but a markscheme would be nice to check if this is what your question wants.
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    Sin(x+pi/6)=0

    x+pi/6= pi*n where n is an integer (this is obvious using the graph of sin i.e you should know sinx is 0 when x=0,pi,2pi,3pi,....)

    x= -pi/6 +pi*n
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    (Original post by 16characterlimit)
    Not 100% sure, but arcsin(0) = 0 +(2pi)n where n is a integer.

    0 + 2(pi)n = x + pi/6

    x = 2pi(n) - pi/6

    The n is because sin is periodic every 2pi radians, this is my interpretation of the answer but a markscheme would be nice to check if this is what your question wants.
    Yeah this isn't true. If this was the case then arcsin would not be a function. It has a restricted range so that it is a function, namely [-\frac{\pi}{2},\frac{\pi}{2}].

    If you're saying what you've highlighted in bold, what you mean is that any number with sine 0 is 0+2n\pi (for some n \in \mathBB{Z}).

    FFS what's wrong with Latex.
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    QUOTE=16characterlimit;65785569]Not 100% sure, but arcsin(0) = 0 +(2pi)n where n is a integer.

    0 + 2(pi)n = x + pi/6

    x = 2pi(n) - pi/6

    The n is because sin is periodic every 2pi radians, this is my interpretation of the answer but a markscheme would be nice to check if this is what your question wants.[/QUOTE]

    Yes, I got x = 2pi(n) -pi/6 but the mark scheme says x = pi(n) - pi/6, I'm not sure if both are accepted?

    I think I could maybe work it out now by considering x + pi/6 as a translation to the left by pi/6. If you look at the graph the answer is computed for every pi(n) - pi/6, but I don't understand how to get the answer if I'm not doing it like that.
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    QUOTE=Math12345;65785619]Sin(x+pi/6)=0

    x+pi/6= pi*n where n is an integer (this is obvious using the graph of sin i.e you should know sinx is 0 when x=0,pi,2pi,3pi,....)

    x= -pi/6 +pi*n[/QUOTE]

    So pi*n is used when sinx is 0, but for any other value 2*pi*n is used?
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    If siny=0 then y=pi*n this should be obvious to you. Then just let y=x+pi/6 and rearrange.
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    (Original post by IrrationalRoot)
    Yeah this isn't true. If this was the case then arcsin would not be a function. It has a restricted range so that it is a function, namely [-\frac{\pi}{2},\frac{\pi}{2}].

    If you're saying what you've highlighted in bold, what you mean is that any number with sine 0 is 0+2n\pi (for some n \in \mathBB{Z}).

    FFS what's wrong with Latex.
    Hehe, correcting me is potentially dangerous.

    OK I removed the offending section.
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    (Original post by Chickenslayer69)
    QUOTE=Math12345;65785619]Sin(x+pi/6)=0

    x+pi/6= pi*n where n is an integer (this is obvious using the graph of sin i.e you should know sinx is 0 when x=0,pi,2pi,3pi,....)

    x= -pi/6 +pi*n
    So pi*n is used when sinx is 0, but for any other value 2*pi*n is used?[/QUOTE]

    Here's the general approach:
    sin(x+pi/6)=0

    x+pi/6=0,pi (2 initial solutions when solving for sin)

    x=-pi/6, x=5pi/6

    x=-pi/6 +2pi*n , x=5pi/6+2pi*n

    Write out a few solutions,

    x=-pi/6, 11pi/6, 5pi/6 , 17pi/6

    Should be obvious general solution is x=-pi/6+pi*n from that.
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    QUOTE=Math12345;65786111]So pi*n is used when sinx is 0, but for any other value 2*pi*n is used?[/QUOTE]

    Here's the general approach:
    sin(x+pi/6)=0

    x+pi/6=0,pi (2 initial solutions when solving for sin)

    x=-pi/6, x=5pi/6

    x=-pi/6 +2pi*n , x=5pi/6+2pi*n

    Write out a few solutions,

    x=-pi/6, 11pi/6, 5pi/6 , 17pi/6

    Should be obvious general solution is x=-pi/6+pi*n from that.[ QUOTE]

    Ohh, ok, got it. Thanks

    I actually did this first but didn't substitute for n so didn't realize it could be simplified to x = -pi/6 + pi*n... I will remember next time.
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    (Original post by Chickenslayer69)
    I can literally do all of them but I can't seem to understand this one??

    http://prntscr.com/bgeogr

    Thanks

    did your answers match up with these?

    UNOFFICIAL MARKSCHEME

    http://www.thestudentroom.co.uk/show....php?t=4168081
 
 
 
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