Hollymae764
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#1
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#1
Can someone check this answer for me:

5sin2x = 3cosx where 0<x<90

i get
10sun(x)cos(x) - 3cos(x) = 0
cos(x) (10sin(x) - 3) = 0

so sinx = 3/10
Last edited by Hollymae764; 1 month ago
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TM3141
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#2
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(Original post by Hollymae764)
Can someone check this answer for me:

5sin3x = 3cosx where 0<x<90

i get
10sun(x)cos(x) - 3cos(x) = 0
cos(x) (10sin(x) - 3) = 0

so sinx = 3/10
That's not right unfortunately. You haven't dealt with the sin3x correctly i dont think
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TM3141
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(Original post by TM3141)
That's not right unfortunately. You haven't dealt with the sin3x correctly i dont think
I also think there must be a typo in this question
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PepeTheFroggi
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#4
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are you sure its not sin2x instead of sin3x?
And if it was sin2x i would divide by cosx to get rid of the cos. Makes solving it easier
Last edited by PepeTheFroggi; 2 months ago
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nerak99
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#5
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well

5sin2x=3cosx becomes 10sinx cosx =3cosx using the double angle formula. However do NOT divide through by cos .

Rather 10sin x cos x - 3cos x=0 >>> cos x(10sinx-3)=0 This gives cos x=0 and sin x =3/10 as solutions which can be evaluated to angles.

If you divide through by cos you would lose the zero root and probably lose an accuracy mark at least.

If you want to aim for A or A* you can't afford to make the mistakes like dividing through by cos x. This is a distinguishing feature of the question and is used to help fix grade boundaries.
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