# Math Help - [SOLVED] Quadratic Trigonometric Equations

1. ## [SOLVED] Quadratic Trigonometric Equations

Hi there, I'm new. 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 $\theta$:

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

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 $\frac {\sqrt {15}}{5}$ and $-\frac {\sqrt {15}}{5}$ . 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.) $tan\theta = 2 sin\theta$

I tried converting to all sin and cos values. I think my mistake was to try to divide both sides by $sin\theta$ after that, since dividing by a variable is a no-no. I ended up with $cos\theta = \frac {1}{2}$ , 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.

2. Originally Posted by Zipo
Hi there, I'm new. 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 $\theta$:

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

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 $\frac {\sqrt {15}}{5}$ and $-\frac {\sqrt {15}}{5}$ . 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.) $tan\theta = 2 sin\theta$

I tried converting to all sin and cos values. I think my mistake was to try to divide both sides by $sin\theta$ after that, since dividing by a variable is a no-no. I ended up with $cos\theta = \frac {1}{2}$ , 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 $\cos 2\theta=1-2\sin^2 \theta$

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

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

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

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

3. Thanks, that helped! In 1.) I was trying to use $sin^2 \theta + cos^2 \theta = 1$ , which didn't help even though it got rid of one of the terms.

So, in 2.) I'm supposed to pull $sin\theta$ 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!