Verifying Solution to 2\sin\theta\cos\theta + 2 \sin\theta - \cos\theta -1 = 0

[Apost8]

Can someone please check over this real quick. I think I’ve solved this problem correctly, but, I’m not sure.

Solve: 2\sin\theta\cos\theta + 2 \sin\theta - \cos\theta -1 = 0


2 \sin\theta(\cos\theta + 1) -(\cos\theta + 1) = 0

2 \sin\theta(\cos\theta + 1) = (\cos\theta + 1)

2 \sin\theta = 1

\sin\theta = 1/2

\theta = (\frac{\pi}{6} + k2\pi)\ \mbox {or}\ (\frac{5\pi}{6} + k2\pi)

 

[topsquark]

Can someone please check over this real quick. I think I’ve solved this problem correctly, but, I’m not sure.

Solve: 2\sin\theta\cos\theta + 2 \sin\theta - \cos\theta -1 = 0


2 \sin\theta(\cos\theta + 1) -(\cos\theta + 1) = 0

2 \sin\theta(\cos\theta + 1) = (\cos\theta + 1)

2 \sin\theta = 1

\sin\theta = 1/2

\theta = (\frac{\pi}{6} + k2\pi)\ \mbox {or}\ (\frac{5\pi}{6} + k2\pi)

Looks good to me. The hard way to check this is to plug your answer into your original equation and do the trig to see if it works. The easy way is to pick several values of k and plug it into your original equation using a calculator.

-Dan

 

[Apost8]

OK, thanks for your help!

 

[VietDao29]

...2 \sin\theta(\cos\theta + 1) -(\cos\theta + 1) = 0

2 \sin\theta(\cos\theta + 1) = (\cos\theta + 1)

2 \sin\theta = 1

Hmm, this step looks wrong.
What if cos \theta + 1 = 0? And you cannot divide both sides by 0!
The better way is to factor out cos \theta + 1, like this:
2 \sin\theta(\cos\theta + 1) -(\cos\theta + 1) = 0
\Leftrightarrow (\cos \theta + 1)(2 \sin\theta - 1) = 0
Now if the product of them is 0, then either \cos \theta + 1 = 0 or 2 \sin\theta - 1 = 0, right?
Can you go from here? :)
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@ topsquark, it's not correct, bud.

 

[topsquark]

Hmm, this step looks wrong.
What if cos \theta + 1 = 0? And you cannot divide both sides by 0!
The better way is to factor out cos \theta + 1, like this:
2 \sin\theta(\cos\theta + 1) -(\cos\theta + 1) = 0
\Leftrightarrow (\cos \theta + 1)(2 \sin\theta - 1) = 0
Now if the product of them is 0, then either \cos \theta + 1 = 0 or 2 \sin\theta - 1 = 0, right?
Can you go from here? :)
--------------
@ topsquark, it's not correct, bud.

Oh, I see your point. That slipped by me. Sorry. Good catch, VeitDao29!

-Dan

 

[Apost8]

Thanks for the help VietDao. Can you please show me how you got \ (\cos \theta + 1)(2 \sin\theta - 1) = 0\? I can’t seem to figure out exactly how you did that.

 

[Apost8]

Oh wait, nevermind, I see how you got that. Duh.

So, from there I would get:

\cos \theta + 1 = 0

\cos \theta = -1

\theta = \pi

OR

2 \sin\theta - 1 = 0

\sin\theta= \frac{1}{2}

\theta = \frac{\pi}{6}, \theta = \frac{5\pi}{6}

 

[Hootenanny]

Looks right to me, what's your limits?

 

[Apost8]

By limits, do you mean, for example (0, 2\pi)?

If that's the case, the question didn't impose any limits, so I suppose the correct answer would be \theta= (\pi+k2\pi,\ \frac{\pi}{6}+k2\pi\ or, \frac{5\pi}{6}+k2\pi).

 

[Hootenanny]

Yeah that's what I meant.

When ever I have solved trig equations I am required to find a quantative answer for \theta, but if your tutor is happy with you leaving it in terms of k that's all that matter.

 

[Apost8]

Thanks for your Help!