# Confused over how to solve trig eqn

• Feb 6th 2011, 05:12 AM
angypangy
Confused over how to solve trig eqn
Hello the equation I want to solve for 0 <= x <= pi

$3 cos^2 x + 2 sin x - 3 = 0$

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
• Feb 6th 2011, 05:16 AM
it's not x=0 or x=2/3 its sinx=0 or sinx=2/3
• Feb 6th 2011, 05:16 AM
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.
• Feb 6th 2011, 05:24 AM
angypangy
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???
• Feb 6th 2011, 05:28 AM
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).
• Feb 6th 2011, 05:32 AM
Unknown008
No, that's not like that.

If:

$\sin(x) = 0$

Then usually:

$x = \sin^{-1}(0) = 0$

And similarly:

$\sin(x) = \dfrac23$

$x = \sin^{-1}\left(\dfrac23\right)= 0.730^c$

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)...
• Feb 6th 2011, 05:45 AM