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Need a little guidance with trig equations. - HSN forum

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Need a little guidance with trig equations.


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#1 Sammi

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Posted 23 December 2006 - 08:27 PM

I have a feeling this is an easy question as we've just started this bit of the trig topic in unit 2.

Solve: cos2x+sinx=0

I know you have to change the cos2x to something else, but that is as far as I've got. smile.gif

#2 Michael

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Posted 23 December 2006 - 09:02 PM

cos2x+sinx=0 you have to use a substitution
-2sin²x + 1 + sinx = 0
(sinx + 2)(sinx - 1) = -1
sinx=-3(not a possible solution) and sinx = 0

Therefore x = 180 and x = 360

#3 Steve

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Posted 24 December 2006 - 01:50 PM

There's a very similar example on page 100 of the notes on Trigonometry.
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#4 ad absurdum

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Posted 24 December 2006 - 01:58 PM

Sorry Micheal that's not quite right.

As Michael said, you replace \cos(2x) with 1-2\sin^2(x) (you really need to know the trig indentities inside out, so you can use them in situations like this without thinking aout it). This gives you:

1-2\sin^2(x) + \sin(x) = 0

Multiply through by -1 and put it in the normal quadratic form to get:

2\sin^2(x) - \sin(x) - 1 = 0

This is just a quadratic equation in sin(x) instead of x. Factorise it in the normal way, it might help to factorise 2x^2 - x - 1 = 0 then replace the x's with sin(x)'s. You should end up with two manageable values of sin(x) that satisfy the quadratic equation. The question should have given you a range (say, 0 \leq x \leq360). All you have to do now is write down the values of x that give a sine of that value in the range. If you weren't given a range in the question, then there are actually infinitely many value of x that satisfy the equation. I'm pretty sure they will always give you a range though.

In general, it's not straightforward to solve equations if they have double angles trig functions in them. I would always convert these into single angle things (that is what Michael done in the first step) then solve from there smile.gif

Edit - Didn't realise Steve had posted there. You should have had a look at the notes before, they are very useful for when you don't know how to approach a general type of problem.
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#5 Michael

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Posted 24 December 2006 - 05:49 PM

Damn I feel embarassed. Higher maths seems so long ago.

#6 st-and Paul

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Posted 24 December 2006 - 07:53 PM

If a range isnt specified better to assume that it will be radians, ie a range of [0, 2*pi] as you dont touch degrees ever again after higher maths. I m in a second year maths uni course and i havent used degrees since higher maths. Radians are easier and more natural to use.

#7 Sammi

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Posted 02 January 2007 - 10:28 PM

Ohh I understand now! I hadn't looked in the notes. I suppose I should have looked there first, thanks guys. smile.gif





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