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Double angle - HSN forum

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Double angle


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#1 Rocky

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Posted 26 April 2006 - 03:18 PM

I can't solve this Q:

(i) Solve the eqn 3sin2x=2sinx for 0 <=x=>360
(ii) Hence state the values of x in the interval 0 <=x=>360 for which 3sin2x<2sinx

#2 Mr H

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Posted 26 April 2006 - 03:48 PM

(a) Rearrange to equal zero.
Use double angle formula for sin2x.
Common factor.
Solve for 'either-or'.

(b) The solutions to (a) are equivalent to the points of intersection of 3sin2x and 2sinx.

Sketch both functions on the same axes and you should be able to write down the intervals where 3sin2x < 2sinx fairly easily.

H tends 2 infinity

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#3 Rocky

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Posted 26 April 2006 - 03:50 PM

'(a) Rearrange to equal zero.
Use double angle formula for sin2x.
Common factor.
Solve for 'either-or'.'

Its 2sinx BTW. Thats why I need help cause its not working for me.

#4 Steve

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Posted 26 April 2006 - 05:05 PM

If you use the identity \sin2x=2\sin x\cos x, then you will get:



\begin{align*}3\sin2x &=2\sin x \\
6\sin{x}\cos{x} &= 2\sin x
\end{align*}

Then you can do the rest of the things that Mr H said.

There are examples in the notes which will probably help.
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#5 Mr H

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Posted 26 April 2006 - 08:14 PM

Rearrange to equal zero.

3sin2x - 2sinx = 0

Use double angle formula for sin2x.

3(2sinxcosx) - 2sinx = 0
6sinxcosx - 2sinx = 0

Common factor.

2sinx(3cosx - 1) = 0

Solve for 'either-or'.

Either 2sinx = 0 implies.gif sinx = 0, etc

Or 3cosx - 1 = 0 implies.gif cosx = 1/3, etc

You could substitute double angle formula first as Steve points out then rearrange, makes no difference.

H tends 2 infinity

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Never argue with an idiot. They drag you down to their level then beat you with experience.






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