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Mathematics 9 Online
OpenStudy (anonymous):

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)

OpenStudy (amistre64):

there is no equation here...

OpenStudy (anonymous):

haha really? why is that?

OpenStudy (amistre64):

Its alot of instructions on what to do; but there is no equation to work it on lol

OpenStudy (anonymous):

lol!! sorry!!

OpenStudy (anonymous):

i got it here

OpenStudy (anonymous):

sintheta + sqrt3costheta = 1

OpenStudy (amistre64):

sin(t) + sqrt(3)cos(t) = 1 right?

OpenStudy (anonymous):

yes

OpenStudy (amistre64):

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

OpenStudy (anonymous):

hahaha thats the way!

OpenStudy (anonymous):

damn it im not gettin it!

OpenStudy (amistre64):

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

OpenStudy (amistre64):

an alternative form is: \[4\sin[\frac{\pi}{4}-\frac{t}{2}] . \sin[\frac{t}{2}+\frac{\pi}{12}]=0\]

OpenStudy (anonymous):

ohkay...lol

OpenStudy (amistre64):

since sin = 0 at 0 and pi; we can prolly figure out a few t's from this ?

OpenStudy (amistre64):

t/2 = pi/4 t = pi/2sounds reasoneable plug it into the orifinal and test

OpenStudy (anonymous):

i think ur getin somewhere haha

OpenStudy (amistre64):

the other seems to be -pi/6 but that aint in the inteval is it lol

OpenStudy (anonymous):

lol no it isnt haha

OpenStudy (amistre64):

sin(pi / 2) + (sqrt(3) * cos(pi / 2)) = 1 is good tho

OpenStudy (anonymous):

would that be the lower value or upper value?

OpenStudy (amistre64):

thats the lower value

OpenStudy (anonymous):

i only get two shots so i probably beta wait till we figur out the upper value lol

OpenStudy (amistre64):

its close to 2pi according to the graph

OpenStudy (amistre64):

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

OpenStudy (amistre64):

11pi/6 right?

OpenStudy (amistre64):

sin((11 * pi) / 6) + (sqrt(3) * cos((11 * pi) / 6)) = 1: from google

OpenStudy (anonymous):

howd u get tht?

OpenStudy (anonymous):

11pi/6 from -p1/6 ?

OpenStudy (amistre64):

all angles have many names to them because when you walk around a circle you can go along way and still get nowhere

OpenStudy (amistre64):

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

OpenStudy (anonymous):

lol. so u added two pi?

OpenStudy (amistre64):

I did ;)

OpenStudy (anonymous):

haha. it works doesnt it?

OpenStudy (amistre64):

it does indeed :) becuase they both have the same values

OpenStudy (anonymous):

kay so u reckon i should give pi/2 and 11pi/6 a shot?

OpenStudy (amistre64):

i do..

OpenStudy (anonymous):

yay ur a genius!!! it works haha

OpenStudy (amistre64):

lol .... i had to remember my trigs :)

OpenStudy (anonymous):

haha. is there a simpler way of doing this tho?

OpenStudy (amistre64):

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

OpenStudy (anonymous):

good ole wolfram lol

OpenStudy (anonymous):

so are you a teacher?

OpenStudy (amistre64):

no, just old :) im trying to get my masters degree in math tho so I can teach college math

OpenStudy (anonymous):

good effort. you'll get there mate haha

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