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# AQA FP1 General Solution Question

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

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2. 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.
3. 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
4. (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 .

If you're saying what you've highlighted in bold, what you mean is that any number with sine 0 is (for some ).

FFS what's wrong with Latex.
5. 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.
6. 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?
7. If siny=0 then y=pi*n this should be obvious to you. Then just let y=x+pi/6 and rearrange.
8. (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 .

If you're saying what you've highlighted in bold, what you mean is that any number with sine 0 is (for some ).

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

OK I removed the offending section.
9. (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.
10. 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.
11. (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

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