# [SOLVED] Quadratic Trigonometric Equations

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• Nov 14th 2009, 09:05 PM
Zipo
[SOLVED] Quadratic Trigonometric Equations
Hi there, I'm new. (Hi) I'd really like to get a tutor's help with this, but I can't wait until Monday, so I hope you won't mind my asking.

Anyway, this has got me really stumped. The book is no help; all the examples are much less complicated. I'll just post two examples I'm having trouble with for now.

In this section I'm supposed to find all values for $\displaystyle \theta$:

1.) $\displaystyle cos (2\theta) + 6 sin^2\theta = 4$

What I tried to do was to move the 4 to the left side, convert to all sine values, and plug the values into the quadratic formula. I ended up with $\displaystyle \frac {\sqrt {15}}{5}$ and $\displaystyle -\frac {\sqrt {15}}{5}$ . This is probably wrong, even though I'm fairly sure I didn't make any algebraic errors (I probably just chose the wrong process).

2.) $\displaystyle tan\theta = 2 sin\theta$

I tried converting to all sin and cos values. I think my mistake was to try to divide both sides by $\displaystyle sin\theta$ after that, since dividing by a variable is a no-no. I ended up with $\displaystyle cos\theta = \frac {1}{2}$ , which resulted in two correct answers (as special angles), but there are supposed to be four.

Any help is greatly appreciated and will result in my abject adoration, and cake.
• Nov 14th 2009, 09:46 PM
mathaddict
Quote:

Originally Posted by Zipo
Hi there, I'm new. (Hi) I'd really like to get a tutor's help with this, but I can't wait until Monday, so I hope you won't mind my asking.

Anyway, this has got me really stumped. The book is no help; all the examples are much less complicated. I'll just post two examples I'm having trouble with for now.

In this section I'm supposed to find all values for $\displaystyle \theta$:

1.) $\displaystyle cos (2\theta) + 6 sin^2\theta = 4$

What I tried to do was to move the 4 to the left side, convert to all sine values, and plug the values into the quadratic formula. I ended up with $\displaystyle \frac {\sqrt {15}}{5}$ and $\displaystyle -\frac {\sqrt {15}}{5}$ . This is probably wrong, even though I'm fairly sure I didn't make any algebraic errors (I probably just chose the wrong process).

2.) $\displaystyle tan\theta = 2 sin\theta$

I tried converting to all sin and cos values. I think my mistake was to try to divide both sides by $\displaystyle sin\theta$ after that, since dividing by a variable is a no-no. I ended up with $\displaystyle cos\theta = \frac {1}{2}$ , which resulted in two correct answers (as special angles), but there are supposed to be four.

Any help is greatly appreciated and will result in my abject adoration, and cake.

For (1) , remember that $\displaystyle \cos 2\theta=1-2\sin^2 \theta$

(2) By dividing both sides by \sin \theta , you are ignoring the fact that $\displaystyle \sin \theta=0$ , and that's why there are 2 missing solutions .

$\displaystyle \frac{\sin \theta}{\cos \theta}=2\sin \theta$

$\displaystyle \sin \theta=2\sin \theta\cos \theta$

$\displaystyle \sin \theta(2\cos \theta-1)=0$
• Nov 15th 2009, 02:12 AM
Zipo
Thanks, that helped! In 1.) I was trying to use $\displaystyle sin^2 \theta + cos^2 \theta = 1$ , which didn't help even though it got rid of one of the terms.

So, in 2.) I'm supposed to pull $\displaystyle sin\theta$ out of the parentheses instead of trying to convert it to a cosine. That makes a lot more sense.

I finished those problems and was able to do the rest of my college homework. Thanks so much!