Homework Help: Trig problem

1. Mar 9, 2006

Apost8

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

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

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

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

$$2 \sin\theta = 1$$

$$\sin\theta = 1/2$$

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

Last edited: Mar 9, 2006
2. Mar 9, 2006

topsquark

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

3. Mar 9, 2006

Apost8

4. Mar 10, 2006

VietDao29

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.

5. Mar 10, 2006

topsquark

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

-Dan

Last edited: Mar 10, 2006
6. Mar 10, 2006

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.

7. Mar 10, 2006

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

$$2 \sin\theta - 1 = 0$$

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

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

Last edited: Mar 10, 2006
8. Mar 10, 2006

Hootenanny

Staff Emeritus
Looks right to me, whats your limits?

9. Mar 10, 2006

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 $$\theta= (\pi+k2\pi,\ \frac{\pi}{6}+k2\pi\ or, \frac{5\pi}{6}+k2\pi)$$.

10. Mar 10, 2006

Hootenanny

Staff Emeritus
Yeah thats 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 $k$ thats all that matter.

11. Mar 10, 2006