Consider the equation 2cos^2x + sinx - 1 = 0
a) Explain why the equation cannot be factored.
b) Suggest a trigonometric identity that can be used to remove the problem identified in part a).
c) Apply the identity and rearrange the equation into a factorable form.
d) Factor the equation.
e) Determine all solutions in the interval x E [0,2pi].
Ok so i'm on part e.. actually factoring it. This is what I come up with:
2cos^2x+sinx-1=0
2(1-sin^2x)+sinx-1=0
2-2sin^2x+sinx-1=0
-2sin^2x+sinx-1+2=0
-2sin^2x+sinx+1=0
So I try to factor it..
(-2x^2+2x)+(-1x+1)
so that looks pretty wrong..
This is what I found on yahoo answers when someone was doing a similar problem:
You know cos²x = 1 - sin²x, so:
2cos²x - sinx - 1 = 0 → 2 (1 - sin²x) - sinx - 1 = 0 →
2 - 2sin²x - sinx - 1 = 0 →
- 2sin²x - sinx + 1 = 0 → 2sin²x + sinx - 1 = 0
How does he go from -2sin^2x to it being positive 2sin^2x, and how come the -sinx all of a sudden switches to +sinx? And the +1 switches to -1? Did he just bring everything on to the other side of the equal sign?
Can anyone clarify if my answers are right?
c) 2cos^2x+sinx-1=0
2(1-sin^2x)+sinx-1=0
2-2sin^2x+sinx-1=0
-2sin^2x+sinx-1+2=0
-2sin^2x+sinx+1=0
2sin^2x-sinx-1=0
d) 2sin^2x-sinx-1=0
and from that I got to
(2sinx+1)(sinx-1)
e) The solutions in the interval [0,2pi] are pi/2, 7pi/6 and 11pi/6.
Is that right?