Question

Can someone help me with this trig question?

solve cos theta>= (sqrt3)/2 for 0<=theta<=2pi

Answer 1

can you draw this or write this in proper mathematical notation?

Answer 2

Uhm I'm not sure what you mean, this is what my problem asked for. I'll try to rewrite it though.

cos θ>/= (squareroot 3)/2 for 0</= θ</= 2pi

Answer 3

by sqrt 3/2, do you mean this\[\sqrt[3]{2}\] or this \[\sqrt{3/2}\]

Answer 4

No, its square root of three separately then put over 2

Answer 5

ok you mean this\[\frac{ \sqrt{3} }{ 2 }\]

Answer 6

Yeah thats it, I donno how to do the symbols here sorry

Answer 7

what is the formula that u can use to solve this ? do u know it ?

Answer 8

No, that's what I don't understand. I used a couple different ones already and they don't work.

Answer 9

when
\[\cos \theta = \ cos\ \alpha \]
\[\theta = 2n \pi \pm \alpha\]
where n is an integer

Answer 10

well \[\cos \theta = \frac{ \sqrt{3} }{ 2 }\] for theta = pi/2 or 90 degrees

Answer 11

that is the equation to solve cos functions
so according to that we know in ur case
\[\cos \theta = \frac{ \sqrt{3} }{ 2}\]
then
\[\cos \theta = \cos 30^{0 }\]
which means
\[\cos \theta = \cos \frac{ \pi }{ 6 }\]
so if
\[\theta = 2n \pi \pm \alpha\]
when
\[\cos \theta = \cos \alpha\]
then here
\[\alpha = \frac{ \sqrt{3} }{ 2 }\]
which means
\[\theta = 2n \pi \pm \alpha\]\[\theta = 2n \pi \pm \frac{ \sqrt{3} }{ 2}\]
now all u have to do is solve it for the given range .Did u got it so far ?

Answer 12

Okay, I think I get it now. This really helps, thanks ^^

Answer 13

u r welcome!!