Need a little guidance with trig equations.
Posted 23 December 2006 - 08:27 PM
I know you have to change the cos2x to something else, but that is as far as I've got.
Posted 23 December 2006 - 09:02 PM
-2sin²x + 1 + sinx = 0
(sinx + 2)(sinx - 1) = -1
sinx=-3(not a possible solution) and sinx = 0
Therefore x = 180 and x = 360
Posted 24 December 2006 - 01:58 PM
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.
Posted 24 December 2006 - 05:49 PM
Posted 24 December 2006 - 07:53 PM
Posted 02 January 2007 - 10:28 PM
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