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.

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# Need a little guidance with trig equations.

Started by Sammi, Dec 23 2006 08:27 PM

6 replies to this topic

### #1

Posted 23 December 2006 - 08:27 PM

### #2

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

-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

Posted 24 December 2006 - 01:50 PM

There's a very similar example on page 100 of the notes on Trigonometry.

### #4

Posted 24 December 2006 - 01:58 PM

Sorry Micheal that's not quite right.

As Michael said, you replace with (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:

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

This is just a quadratic equation in sin(x) instead of x. Factorise it in the normal way, it might help to factorise 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, ). 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

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.

As Michael said, you replace with (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:

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

This is just a quadratic equation in sin(x) instead of x. Factorise it in the normal way, it might help to factorise 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, ). 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

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

Posted 24 December 2006 - 05:49 PM

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

### #6

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

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.

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