# Math Help - Consider the equation 2cos^2x+sinx-1=0

1. ## Consider the equation 2cos^2x+sinx-1=0

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].

2. 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

3. 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 $x$ 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.

4. 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 $\cos{x}$ and $\sin{x}$ with the same variable.

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

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

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

6. 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?

7. 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)

8. 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?

9. 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.