Solve this equation on the interval 0\[\le\] θ < 2π. If there is only one solution, write it in the upper box and write NONE in the lower box. If there is no solution, write NONE in both boxes. Round your answer(s) to two decimal places using radians. θ = (smaller value of θ) θ = (larger value of θ)

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Solve this equation on the interval 0\[\le\] θ < 2π. If there is only one solution, write it in the upper box and write NONE in the lower box. If there is no solution, write NONE in both boxes. Round your answer(s) to two decimal places using radians. θ = (smaller value of θ) θ = (larger value of θ)

Mathematics
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there is no equation here...
haha really? why is that?
Its alot of instructions on what to do; but there is no equation to work it on lol

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Other answers:

lol!! sorry!!
i got it here
sintheta + sqrt3costheta = 1
sin(t) + sqrt(3)cos(t) = 1 right?
yes
then.... sqrt(3)cos(t) = 1 - sin(t) 3cos^2 = 1 +sin^2 -2sin... lets cheat and do this lol http://www.wolframalpha.com/input/?i=sin%28t%29+%2B+sqrt%283%29cos%28t%29+%3D+1
hahaha thats the way!
damn it im not gettin it!
i cant seem to get it out either.... i know theres one answer at least; might have to just make a chart and narrow it in
http://www.wolframalpha.com/input/?i=Solve%5Bsin%28t%29+%2B+sqrt%283%29cos%28t%29+%3D%3D+1%2Ct%5D
an alternative form is: \[4\sin[\frac{\pi}{4}-\frac{t}{2}] . \sin[\frac{t}{2}+\frac{\pi}{12}]=0\]
ohkay...lol
since sin = 0 at 0 and pi; we can prolly figure out a few t's from this ?
t/2 = pi/4 t = pi/2sounds reasoneable plug it into the orifinal and test
i think ur getin somewhere haha
the other seems to be -pi/6 but that aint in the inteval is it lol
lol no it isnt haha
sin(pi / 2) + (sqrt(3) * cos(pi / 2)) = 1 is good tho
would that be the lower value or upper value?
thats the lower value
i only get two shots so i probably beta wait till we figur out the upper value lol
its close to 2pi according to the graph
now we know -pi/6 is good; but if we take and try to reach that going from 0 to 2pi what do we get? i wonder
11pi/6 right?
sin((11 * pi) / 6) + (sqrt(3) * cos((11 * pi) / 6)) = 1: from google
howd u get tht?
11pi/6 from -p1/6 ?
all angles have many names to them because when you walk around a circle you can go along way and still get nowhere
we can walk -pi/6 to smell the roses; or we can take the long way around; and walk 11pi/6 to get to the same place to wmell the roses
lol. so u added two pi?
I did ;)
haha. it works doesnt it?
it does indeed :) becuase they both have the same values
kay so u reckon i should give pi/2 and 11pi/6 a shot?
i do..
yay ur a genius!!! it works haha
lol .... i had to remember my trigs :)
haha. is there a simpler way of doing this tho?
i caouldnt find a way to get it into the alternate form; but wolfram did... so i tested that forms results into the original and they worked like a charm
good ole wolfram lol
so are you a teacher?
no, just old :) im trying to get my masters degree in math tho so I can teach college math
good effort. you'll get there mate haha

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