Preliminary Question: How do i solve this?? (1 Viewer)

[foram]

HELP HELP!!!

2cos2@ = tan(5pi/4) for 0<@<2pi

Can anybody solve this? I have no idea how its done.

@ is theta (an angle) in this example

 

[negativerootof2]

Re: How do i solve this??

not sure but if you evaluate 5pi/4 it comes to 1

so 2cos2@ = 1
cos2@ = 1/2

using table of exact values thing, maybe another method
cos pi/3 = 1/2

therefore 2@ = pi/3
and @ = pi/6

and I can't remember how you find all the other values for @ in the circle

 

[bored of sc]

Re: How do i solve this??

do you somehow get @ on its own (solving an equation) ?? i'm confused also

 

[foram]

Re: How do i solve this??

wow, tan 5pi/4 is actually 1, i missed that part, now it's only a simple 2cos2@=1 equation! thanks for the help so much!!

 

[negativerootof2]

Re: How do i solve this??

no worries XD

 

[tacogym27101990]

yeah doubt its a question 5 or 6 in ext 1

 

[Mark576]

Re: How do i solve this??

not sure but if you evaluate 5pi/4 it comes to 1

so 2cos2@ = 1
cos2@ = 1/2

using table of exact values thing, maybe another method
cos pi/3 = 1/2

therefore 2@ = pi/3
and @ = pi/6

and I can't remember how you find all the other values for @ in the circle

cos2@=1/2
2@=pi/3, 5pi/3, 7pi/3, 11pi/3 [We adjust the domain for 2@: it becomes 0&le;2@&le;4pi]
Then:
@=pi/6, 5pi/6, 7pi/6, 11pi/6

Notice I found the other answers by considering where the cosine function is positive in the unit circle, which is the first and fourth quadrants. So the other angle in a single revolution will be given by 2pi - @, where @ is the acute angle.