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].

Originally Posted by kmjt

Ok.. so I tried to factor it. This is my steps:

2cos^2x + sinx - 1 =0

Err i change the way it looks to make it easier for me

2x^2 + x - 1 =0

2x^2 + 2x -1x -1

(2x^2 + 2x) + (-1x - 1)

2x(x+1) -1(x+1)

(2x-1)(x+1)

(2cosx - 1)(cosx + 1)

Isn't that factoring it? Why is it asking why you cannot factor it? What am I doing wrong?

hmm... A silly mistake

The equation is 2cos^2x + sinx - 1 = 0 and not 2cos^2x + cosx - 1 =0

Originally Posted by kmjt

Ok.. so I tried to factor it. This is my steps:

2cos^2x + sinx - 1 =0

Err i change the way it looks to make it easier for me

2x^2 + x - 1 =0

...

Isn't that factoring it? Why is it asking why you cannot factor it? What am I doing wrong?

It looks like you've substituted $\displaystyle x$ for both cos x and sin x. That's not good.

The way to proceed is to use a suitable trigonometric identity ... say one involving squares of cosines and squares of sines, somehow ... to transform your cos squared term into a sin squared term.

Originally Posted by kmjt

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 tried to factor it. This is my steps:

2cos^2x + sinx - 1 =0

Err i change the way it looks to make it easier for me

2x^2 + x - 1 =0

2x^2 + 2x -1x -1

(2x^2 + 2x) + (-1x - 1)

2x(x+1) -1(x+1)

(2x-1)(x+1)

(2cosx - 1)(cosx + 1)

Isn't that factoring it? Why is it asking why you cannot factor it? What am I doing wrong?

you cannot represent $\displaystyle \cos{x}$ and $\displaystyle \sin{x}$ with the same variable.


$\displaystyle 2\cos^2{x} + \sin{x} - 1 = 0
$

change $\displaystyle cos^2{x}$ to $\displaystyle 1 - \sin^2{x}$ and try again.

I see! I'm blind I didn't notice that it was sin :$ Thanks guys.

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?

Originally Posted by kmjt

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?

yes ... or you can look at it as multiplying both sides by (-1)

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?

Originally Posted by kmjt

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?

Yes.