Student20
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http://filestore.aqa.org.uk/subjects...W-QP-JAN09.PDF

Q 2B, ii)

I've got sqrt(10)sin(x-1.25) = -sqrt(10)

-sqrt(10) / sqrt(10) = -1


sin(x-1.25) = -1

sin^-1(-1) = -1/2(pi)

-1/2pi = x - 1.25

x = -1/2pi + 1.25

The actual answer is 5.96 confused as to how they got there

Any help appreciated
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Kevin De Bruyne
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(Original post by Student20)
http://filestore.aqa.org.uk/subjects...W-QP-JAN09.PDF

Q 2B, ii)

I've got sqrt(10)sin(x-1.25) = -sqrt(10)

-sqrt(10) / sqrt(10) = -1


sin(x-1.25) = -1

sin^-1(-1) = -1/2(pi)

-1/2pi = x - 1.25

x = -1/2pi + 1.25

The actual answer is 5.96 confused as to how they got there

Any help appreciated
The range in the question may help
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Student20
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0<x<pi/2

Not sure really how the range affects it
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Student20
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Anyone?
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Zacken
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(Original post by Student20)
Anyone?
Have you heard of the CAST diagram/method?

You have \sin (x-1.25) = -1.

So your "key-angle" or "basic-angle" is \sin^{-1} |-1| = \sin^{-1} (1) = \frac{\pi}{2}.

Then because your sine is "negative", then your solutions are in the 'third' or 'fourth quadrant.

So  x-1.25 = \pi - \frac{\pi}{2} or x-1.25 = 2\pi - \frac{\pi}{2}.

Solve both of those equations for x and you should get your answer.
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