Solving Trigonometry Equations

[johnstobbart]

Homework Statement

My trig is really rusty and I've been trying to figure out why the answer is what it is:
Find all solutions of the following equation:
sin² x + cos x - 1.

Homework Equations

Just the identity:
sin² θ + cos² θ = 1

The Attempt at a Solution

The sin² x is in the way, so I substitute 1 - cos² x in it's place to get:
(1 - cos² x + cos x - 1 = 0

The ones are removed and cos x is common, so:
cos x(-cos x + 1) = 0

This means that:
cos x = 0 or cos x = -1

From the unit circle, we get:

cos x = ∏/2 or cos = ∏

Here is where I get confused:

I know we need to make this true for all intervals so:

cos x = 0 whenever x = ± ∏/2 + 2k∏ or,
cos x = -1 whenever x = ± ∏ + 2k∏ for any integer k.

That's my final answer. According to my textbook, It's wrong, but I have no idea why. The textbook gives the final answer as:

cos x = 0 whenever x = ∏/2 + 2k∏ or,
cos x = -1 whenever x = ∏ + 2k∏ for any integer k.

Is it wrong that x = ± ∏/2 or x = ±∏?

 

[ehild]

Homework Statement

My trig is really rusty and I've been trying to figure out why the answer is what it is:
Find all solutions of the following equation:
sin² x + cos x - 1.

Homework Equations

Just the identity:
sin² θ + cos² θ = 1

The Attempt at a Solution

The sin² x is in the way, so I substitute 1 - cos² x in it's place to get:
(1 - cos² x + cos x - 1 = 0

The ones are removed and cos x is common, so:
cos x(-cos x + 1) = 0

This means that:
cos x = 0 or cos x = -1

From the unit circle, we get:

cos x = ∏/2 or cos = ∏

Here is where I get confused:

I know we need to make this true for all intervals so:

cos x = 0 whenever x = ± ∏/2 + 2k∏ or,
cos x = -1 whenever x = ± ∏ + 2k∏ for any integer k.

That's my final answer. According to my textbook, It's wrong, but I have no idea why. The textbook gives the final answer as:

cos x = 0 whenever x = ∏/2 + 2k∏ or,
cos x = -1 whenever x = ∏ + 2k∏ for any integer k.

Is it wrong that x = ± ∏/2 or x = ±∏?

Check the piece in red.

By the way, x = ± ∏ + 2k∏ means the same x values as x = ∏ + 2k∏.

If you take n=k-1 then - ∏ + 2k∏=∏+2(k-1)∏=∏+2n∏.ehild

 

[johnstobbart]

Thanks for the correction.
I see now. It's not wrong, just redundant.

Thanks a lot for your help.

 

[ehild]

Yes, but the solution is not correct. -cosx+1=0 does not mean cosx=-1.

ehild

 

[johnstobbart]

Yes. - cos x + 1 = 0 is supposed to be cos x = 1, right?

 

[ehild]

Yes. - cos x + 1 = 0 is supposed to be cos x = 1, right?

Yes. So x=?

ehild

 

[johnstobbart]

Sorry for the late reply. Things have been hectic.

x = 0, I think.

 

[Mentallic]

Sorry for the late reply. Things have been hectic.

x = 0, I think.

Come on, you're learning to provide all the solutions here! What else does x equal?

 

[johnstobbart]

x = 0 whenever x = ∏/2 + 2∏k or
x = 1 whenever x = 2∏ + 2∏k for any integer k.

Was there a mistake in the textbook when it said cos x = -1 whenever x = ∏ + 2k∏ for any integer k.

 

[Mentallic]

x = 0 whenever x = ∏/2 + 2∏k or
x = 1 whenever x = 2∏ + 2∏k for any integer k.

Was there a mistake in the textbook when it said cos x = -1 whenever x = ∏ + 2k∏ for any integer k.

I believe you were meant to say
$\cos(x)=0$ whenever x=...
as opposed to x=0.

Now, you're nearly correct. Take a look at the graph of $y=\cos(x)$ and you should notice that while it touches its extreme values of -1 and 1 only once every $2\pi$ units, it cuts everything else in between twice every $2\pi$ units. It cuts the y-axis (or cos(x)=0) every $\pi$ units, so your answer should be

$\cos(x)=0$ whenever $x=\frac{\pi}{2}+\pi k$ which you can also represent (and should become familiar with) the factorized form $x=\pi(\frac{1}{2}+k)$

Also, for $\cos(x)=1$, your answer of $x=2\pi+2\pi k$ is a convoluted way of saying $x=2\pi k$ which is the same thing. Do you see why?

 

[johnstobbart]

Yes, I did mean cos (x) = 0 whenever x = ...

I see it now. Trig has never been my strong point and I'll make sure to practise it some more, especially the basics.

Thanks a whole lot for being so patient Mentallic and ehild.

 

[johnstobbart]

Yes, I did mean cos (x) = 0 whenever x = ...

I see it now. Trig has never been my strong point and I'll make sure to practise it some more, especially the basics.

Thanks a whole lot for being so patient Mentallic and ehild.