Question

Find all solutions in the interval 0 12 years ago

Answer 1

I first divide the right by 6 making it 1/2 then take the square root of both sides getting me cos(theta)= sqrt(2)/2....

Answer 2

but then thats when i get confused

Answer 3

hi,
\[6 \cos ^{2} \theta = 3\]
\[\cos ^{2}\theta = \frac{ 3 }{6 } = \frac{ 1 }{2 }\]
\[\cos \theta = \pm \sqrt{\frac{ 1 }{ 2 }}\]
\[\cos \theta = \pm \frac{ 1 }{\sqrt{2} }\]

when \[\cos \theta = \frac{ 1 }{ \sqrt{2} }\]
\[\cos \theta = \cos \frac{ \pi }{ 4 }\]
\[\theta = 2n \pi \pm \frac{ \pi }{4 } where \ \left( n \epsilon \mathbb{Z} \right)\]

when \[\cos \theta = - \frac{ 1 }{ \sqrt{2} }\]
\[\cos \theta = \cos( \pi - \frac{ \pi }{ 4 })\]
\[\theta = 2n \pi \pm ( \pi - \frac{ \pi }{ 4 }) where \ \left( n \epsilon \mathbb{Z} \right)\]

Answer 4

did ya get it... ? srry it takes time to type the equations :/

Answer 5

thank you for your detailed explanation.. but isnt 1/sqrt2 just sqrt2/2?...

Answer 6

I've always understood that you can never have a sqrt at the bottom. maybe im wrong

Answer 7

yeah!! they r same....

Answer 8

ahhh ok, big big help thanks man!!!! i really appreciate it.

Answer 9

Rvenegas13=Fan lmfao

Answer 10

but don't u know that
\[\cos \frac{ \pi }{ 4} = \frac{ 1 }{ \sqrt{2} } = \frac{ \sqrt{2} }{2 }\]

Answer 11

which is 45deg

Answer 12

yep!!