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Hello the equation I want to solve for 0 <= x <= pi
This is what I have so far:
3(1-sin^2 x) + 2 sin x - 3 = 0
3 - 3sin^2 x + 2 sin x - 3 = 0
-3 sin^2 x + 2 sin x = 0
sin x (2 - 3sin x) = 0
roots: x = 0 or 2/3
Is this correct? But now what? Do I calculate cos(0)? and cos(2/3)?
Angus
it's not x=0 or x=2/3 its sinx=0 or sinx=2/3
You haven't completed your work to a solution.
Your final line is incorrect.
What you have are the possible values for sinx.
What about x ?
That's what you are being asked to solve for.
OK it's not x=0 or x=2/3 its sinx=0 or sinx=2/3
but sin(0) = 0 and sin(2/3) = 0.72972765622696636345479665981332. So are they the answers?
I have to express in terms of pi. So 0 for first solution. But 0.72972765622696636345479665981332???
That's x in radians.
There are 2 more answers
(think of sinx as being the vertical co-ordinate of a point on the unit circle).
No, that's not like that.
If:
Then usually:
And similarly:
Usually though, you need to find the critical angle, and from then, you get the x that lie within the boundaries imposed to you, because in 2pi rad, you have two solutions for one sine ratio and in pi, you only have one solution most of the time. Here is one of those exceptions.
EDIT: I thought you did sin(2/3). Note that it's sin^-1(2/3) or arcsine(2/3)...
Here are some visuals
Angus, don't forget to try finding all 4 solutions for x.
The Cosine is unimportant, you just need to find all x that make the original equation zero.