Solve 3cos2θ + 5cosθ + 2 = 0 for 0° ≤ θ

Question

Solve 3cos2θ + 5cosθ + 2 = 0 for 0° ≤ θ < 180°

anonymous · Answer

Do you know the trigonometric identity for double angles?   
You should begin by changing cos2θ so that you have only cosθ to work with.

ttop0816 · Answer

Ø
{109°} <<< this one?
{132°}

anonymous · Answer

Cos2θ = cos^2θ - sin^2θ

You can change this so that you are only dealing with cosθ.

ttop0816 · Answer

how would i go with the equation there??

anonymous · Answer

Do you know any identities?  You must if you've been given this question.
sin^2θ = 1-cos^2θ should help.

tkhunny · Answer

Is it really $\cos(2\theta)$ or did you intend $\cos^{2}(\theta)$?

ttop0816 · Answer

??

anonymous · Answer

lol latex fail
days now

anonymous · Answer

the question is, is it cosine of two x, or cosine squared of x
cos(2x) or cos^2(x)?

anonymous · Answer

i am going to guess it is 3cos^2(x)+5cos(x)+2=0
because this one factors as 
(cos(x) + 1)(3cos(x) + 2)=0
that gives you 
cos(x) = -1, or cos(x) = -2/3

anonymous · Answer

cos(x) = -1 tells you x = 180

anonymous · Answer

cos(x) = -2/3 you need a calculator to solve
find arccos(-2/3)

ttop0816 · Answer

132 degrees!

anonymous · Answer

@KinzaN fyi "latex" is the math writing tool here, and it is not working for the last few days
that is why we have to write by hand -2/3 instead of $-\frac{2}{3}$

anonymous · Answer

yes, 132 rounded

tkhunny · Answer

I keep hearing about LaTeX failures, but I have yet to see one on this round.  I am using FireFox.  Maybe it's an IE thing?