# Trigonomic Equations

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• Aug 19th 2011, 02:00 PM
UnstoppableBeast
Trigonomic Equations
$cos^2(x)-cos2(x)+sin(x)=-sin^2(x)$

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 $cos2(x)=cos^2(x)-sin^2(x)$ I reached: $-sin^2(x)+sin(x)=-sin^2(x)$ So...can that even be solved like that or have I done something wrong?

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

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? (Happy)
• Aug 19th 2011, 02:19 PM
pickslides
Re: Trigonomic Equations
$\cos^2(x)-\cos(2x)+\sin(x)=-\sin^2(x)$

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

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

What happens next?
• Aug 19th 2011, 02:22 PM
Quacky
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:

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

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

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

Let $t=sin(x)$

$2t^2+t=0$

How would you solve that?

Edit: Whoops, beaten to it, sorry!
• Aug 19th 2011, 02:24 PM
UnstoppableBeast
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?
• Aug 19th 2011, 02:26 PM
pickslides
Re: Trigonomic Equations
Quote:

Originally Posted by Quacky

$2t^2+t=0$

How would you solve that?

...
• Aug 19th 2011, 03:46 PM
UnstoppableBeast
Re: Trigonomic Equations
$2t^2=-t$

$2t=-1$

$t=-1/2$

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.
• Aug 19th 2011, 03:54 PM
Quacky
Re: Trigonomic Equations
Quote:

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

$2t=-1$

$t=-1/2$

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.

$2t^2+t=0$

Factoring:

$t(2t+1)=0$

$t=0$ or $t=-0.5$

But you're not done. You know what $t$ is. So what is $x$?
• Aug 19th 2011, 04:02 PM
UnstoppableBeast
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.

$sin(x)=0$

$sin(x)=-.5$

$x= 0, 11pi/6$
• Aug 19th 2011, 04:39 PM
Siron
Re: Trigonomic Equations
Don't forget there are infinite solutions.
• Aug 19th 2011, 04:49 PM
UnstoppableBeast
Re: Trigonomic Equations
Thanks for your help everyone :3
• Aug 19th 2011, 06:29 PM
Prove It
Re: Trigonomic Equations
Quote:

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

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 $cos2(x)=cos^2(x)-sin^2(x)$ I reached: $-sin^2(x)+sin(x)=-sin^2(x)$ So...can that even be solved like that or have I done something wrong?

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

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? (Happy)

2.
\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 $\displaystyle \csc{x} = -1$ and $\displaystyle \csc{x} = 2$
• Aug 19th 2011, 06:31 PM
terrorsquid
Re: Trigonomic Equations
Quote:

Originally Posted by Siron
Don't forget there are infinite solutions.

So would it be:

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

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

$x = \frac{11\pi}{6} + 2\pi k$ where $k \in \mathbb{Z}$
• Aug 19th 2011, 10:33 PM
Siron
Re: Trigonomic Equations
@ terrorsquid: You forgot one solution!

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

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

So four different solutions:
$V=\{2k\pi, \pi+2k\pi, \frac{-\pi}{6}+2k\pi, \frac{7\pi}{6}+2k\pi \}$
( $k\in \mathbb{Z}$)
• Aug 20th 2011, 01:41 AM
terrorsquid
Re: Trigonomic Equations
Isn't $\{\pi k\}$ the same as $\{2k\pi, \pi+2k\pi\}$ because starting from and including 0 + any integer multiple of $\pi$ would = $\pi$ or 0?
• Aug 20th 2011, 01:43 AM
Siron
Re: Trigonomic Equations
To check we write down a few solution of:
$\{k\pi \}=-\pi,0,\pi,2\pi,3\pi, ...$
$\{2k\pi\}=-2\pi,0,2\pi,4\pi,...$
$\{\pi +2k\pi\}=-\pi,\pi, 3\pi, 5\pi, ...$

So indeed that's an option, my apologies.
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