Question

What values for theta (0<=theta<=2pi) satisfy the equation 2 cos theta+1=-cos theta?

Answer 1

\[ \theta (0\le \theta \le 2 \pi ) \\
2 \cos \theta+1=-\cos \theta\]

Answer 2

\(\bf 2 cos (\theta)+1=-cos(\theta) \ \ ?\)

Answer 3

here are the choices

Answer 4

i don't really know where to start on this one at all

Answer 5

is it => \(\bf \bf 2 cos (\theta+1)=-cos(\theta) \ \ ?\) or \(\bf \bf 2 cos (\theta)+1=-cos(\theta) \ \ ?\)

Answer 6

no its the second

Answer 7

or wait
no i think you're right i misread that

Answer 8

it is 2 cos theta+1

Answer 9

hmmm no, I think you're right, is \(\bf 2 cos (\theta)+1=-cos(\theta)\)

Answer 10

\(\bf 2 cos (\theta)+1=-cos(\theta) \implies
3 cos (\theta)+1 = 0 \implies cos (\theta) = \\
cos^{-1}(cos (\theta)) = cos^{-1}\left(-\cfrac{1}{3}\right) \implies \theta = cos^{-1}\left(-\cfrac{1}{3}\right)\)

then just get that, arcCosine in Radians, as far as I can tell is the choices

Answer 11

hmmm

Answer 12

\(\bf 2 cos (\theta)+1=-cos(\theta) \implies
3 cos (\theta)+1 = 0 \implies cos (\theta) = -\cfrac{1}{3}\\
cos^{-1}(cos (\theta)) = cos^{-1}\left(-\cfrac{1}{3}\right) \implies \theta = cos^{-1}\left(-\cfrac{1}{3}\right)\)

Answer 13

so is it 1.91 and 4.37?

Answer 14

well, what did you get?

Answer 15

ohh, I see, you got 1.91

Answer 16

is that right

Answer 17

yes, that's it, see the interval is \(\bf \theta \in (0\le \theta \le 2 \pi )\)
0 = 0 Radians
\(\bf 2\pi = 2\times 3.1416 \implies 6.28\)
so 1.91 and 4.37 Radians are well within that rage

Answer 18

range rather

Answer 19

yay thank you, i usually always guess but now I kind of understand