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# FP1: General Solution, not in MS Watch

1. QP:
http://filestore.aqa.org.uk/subjects...W-QP-JAN11.PDF

MS:
http://filestore.aqa.org.uk/subjects...W-MS-JAN11.PDF

Question 4.

This is my working out.
Inverse sin, got -pi/6.

Therefore.
In my book it says: x = 2piN + (pi - alpha).

4x - 2pi/3 = 2piN + pi + pi/6.
4x = 2piN + 11pi/6.
x = .5piN + 11pi/24

So why is this answer not in the mark scheme? Have I made a mistake somewhere?
2. (Original post by Konnichiwa)
QP:
http://filestore.aqa.org.uk/subjects...W-QP-JAN11.PDF

MS:
http://filestore.aqa.org.uk/subjects...W-MS-JAN11.PDF

Question 4.

This is my working out.
Inverse sin, got -pi/6.

Therefore.
In my book it says: x = 2piN + (pi - alpha).

4x - 2pi/3 = 2piN + pi + pi/6.
4x = 2piN + 11pi/6.
x = .5piN + 11pi/24

So why is this answer not in the mark scheme? Have I made a mistake somewhere?
Generally speaking you get two possible values of theta in the range from 0 to 2pi when you solve sin(theta) = some value.

You can add 2pi to each of these values to produce a general solution. There is a way to combine them, but I prefer to answer with two families.
3. I'm not sure what formula you're quoting, but the way I do it is:
, so and so gives pi/8 + pi n/2.

Also , so we get , and we can repeat the previous steps.

(I misread the question the first time…)

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