1. ## Trigonomic Equations

$\displaystyle cos^2(x)-cos2(x)+sin(x)=-sin^2(x)$

There's the problem, I've tried various different solutions but I can't seem to get it right, I've noticed that the middle cosine is a double-angle but I can't seem to make the formula give me an actual answer.

For example, after canceling out and using the double-angle formula of $\displaystyle cos2(x)=cos^2(x)-sin^2(x)$ I reached: $\displaystyle -sin^2(x)+sin(x)=-sin^2(x)$ So...can that even be solved like that or have I done something wrong?

$\displaystyle csc^2(x)-csc(x)=2$

Same as the one above, despite the fact that it looks simpler, I'm basically incompetent. I have a slew of other problems similar to these and I think some examples will really help me figure out the others. Maybe there's some basic formula/identity that I'm missing out on.

Note: I'm sorry for the title, I couldn't think of anything more specific. At least I'm following all the rules this time, right?

2. ## Re: Trigonomic Equations

$\displaystyle \cos^2(x)-\cos(2x)+\sin(x)=-\sin^2(x)$

$\displaystyle \cos^2(x)+\sin^2(x)-\cos(2x)+\sin(x)=0$

$\displaystyle \cos^2(x)+\sin^2(x)-(\cos^2(x)-\sin^2(x))+\sin(x)=0$

What happens next?

3. ## Re: Trigonomic Equations

You're on the right lines but I think I noticed a sign error with the original post. Using the identitity you stated, we get:

$\displaystyle cos^2(x)-[cos^2(x)-sin^2(x)]+sin(x)=-sin^2(x)$
Notice the signs? Check them below:
$\displaystyle cos^2(x)-cos^2(x)+sin^2(x)+sin(x)=-sin^2(x)$

$\displaystyle sin^2(x)+sin(x)=-sin^2(x)$

$\displaystyle 2sin^2(x)+sin(x)=0$

Let $\displaystyle t=sin(x)$

$\displaystyle 2t^2+t=0$

How would you solve that?

Edit: Whoops, beaten to it, sorry!

4. ## Re: Trigonomic Equations

Ah I see, the initial part of the equation becomes "1" due to the pythagoreon identities but what I don't understand is what happens after. How do I get an actual answer in radians/degrees?

5. ## Re: Trigonomic Equations

Originally Posted by Quacky

$\displaystyle 2t^2+t=0$

How would you solve that?

...

6. ## Re: Trigonomic Equations

$\displaystyle 2t^2=-t$

$\displaystyle 2t=-1$

$\displaystyle t=-1/2$

Thanks for the help, it was a brain fart I suppose.

Edit- Changing the sin/cos/tan into variables like "t" or "x" helped a lot, it's that kind of thinking/problem solving skill that I should've thought of. Often simplifying things makes it easier for my mind (and I assume everyone else's) to comprehend.

7. ## Re: Trigonomic Equations

Originally Posted by UnstoppableBeast
$\displaystyle 2t^2=-t$

$\displaystyle 2t=-1$

$\displaystyle t=-1/2$

Thanks for the help, it was a brain fart I suppose.
Ahhh! You need to do a quick review on algebra. Never divide by a variable - you lose a solution.

$\displaystyle 2t^2+t=0$

Factoring:

$\displaystyle t(2t+1)=0$

$\displaystyle t=0$ or $\displaystyle t=-0.5$

But you're not done. You know what $\displaystyle t$ is. So what is $\displaystyle x$?

8. ## Re: Trigonomic Equations

O_O Wtf is wrong with me, I'm supposed to be going into an AP Calculus class as a 4.0 Student this year and I've forgotten everything.

Anyways, I can handle the rest from here.

$\displaystyle sin(x)=0$

$\displaystyle sin(x)=-.5$

$\displaystyle x= 0, 11pi/6$

9. ## Re: Trigonomic Equations

Don't forget there are infinite solutions.

10. ## Re: Trigonomic Equations

Thanks for your help everyone :3

11. ## Re: Trigonomic Equations

Originally Posted by UnstoppableBeast
$\displaystyle cos^2(x)-cos2(x)+sin(x)=-sin^2(x)$

There's the problem, I've tried various different solutions but I can't seem to get it right, I've noticed that the middle cosine is a double-angle but I can't seem to make the formula give me an actual answer.

For example, after canceling out and using the double-angle formula of $\displaystyle cos2(x)=cos^2(x)-sin^2(x)$ I reached: $\displaystyle -sin^2(x)+sin(x)=-sin^2(x)$ So...can that even be solved like that or have I done something wrong?

$\displaystyle csc^2(x)-csc(x)=2$

Same as the one above, despite the fact that it looks simpler, I'm basically incompetent. I have a slew of other problems similar to these and I think some examples will really help me figure out the others. Maybe there's some basic formula/identity that I'm missing out on.

Note: I'm sorry for the title, I couldn't think of anything more specific. At least I'm following all the rules this time, right?
2.
\displaystyle \displaystyle \begin{align*} \csc^2{x} - \csc{x} &= 2 \\ \csc^2{x} - \csc{x} - 2 &= 0 \\ X^2 - X - 2 &= 0 \textrm{ after making the substitution }X = \csc{x} \\ (X + 1)(X - 2) &= 0 \\ X + 1 = 0\textrm{ or }X - 2 &= 0 \\ X = -1\textrm{ or }X &=2 \end{align*}

So solve $\displaystyle \displaystyle \csc{x} = -1$ and $\displaystyle \displaystyle \csc{x} = 2$

12. ## Re: Trigonomic Equations

Originally Posted by Siron
Don't forget there are infinite solutions.
So would it be:

$\displaystyle x = \pi k$ where $\displaystyle k \in \mathbb{Z}$

$\displaystyle x = \frac{7\pi}{6} + 2\pi k$ where $\displaystyle k \in \mathbb{Z}$

$\displaystyle x = \frac{11\pi}{6} + 2\pi k$ where $\displaystyle k \in \mathbb{Z}$

13. ## Re: Trigonomic Equations

@ terrorsquid: You forgot one solution!

Solve $\displaystyle \cos^2(x)-\cos(2x)+\sin(x)=-\sin^2(x)$
$\displaystyle \Leftrightarrow \cos^2(x)-\cos(2x)+\sin(x)+\sin^2(x)=0$
$\displaystyle \Leftightarrow 1-\cos(2x)+\sin(x)=0$
$\displaystyle \Leftrightarrow 1-[1-2\sin^2(x)]+\sin(x)=0$
$\displaystyle \Leftrightarrow 2\sin^2(x)+\sin(x)=0$
$\displaystyle \Leftrightarrow \sin(x)[2\sin(x)+1]=0$

Two different cases:
$\displaystyle \sin(x)=0 \Leftrightarrow x=2k\pi \ \mbox{or} \ x=\pi + 2k\pi$
$\displaystyle \sin(x)=\frac{-1}{2} \Leftrightarrow x=-\frac{\pi}{6}+2k\pi \ \mbox{or} \ x=\pi+\frac{\pi}{6}+2k\pi$

So four different solutions:
$\displaystyle V=\{2k\pi, \pi+2k\pi, \frac{-\pi}{6}+2k\pi, \frac{7\pi}{6}+2k\pi \}$
($\displaystyle k\in \mathbb{Z}$)

14. ## Re: Trigonomic Equations

Isn't $\displaystyle \{\pi k\}$ the same as $\displaystyle \{2k\pi, \pi+2k\pi\}$ because starting from and including 0 + any integer multiple of $\displaystyle \pi$ would = $\displaystyle \pi$ or 0?

15. ## Re: Trigonomic Equations

To check we write down a few solution of:
$\displaystyle \{k\pi \}=-\pi,0,\pi,2\pi,3\pi, ...$
$\displaystyle \{2k\pi\}=-2\pi,0,2\pi,4\pi,...$
$\displaystyle \{\pi +2k\pi\}=-\pi,\pi, 3\pi, 5\pi, ...$

So indeed that's an option, my apologies.

Page 1 of 2 12 Last