anonymous
  • anonymous
solve for x: cos(x) - sin(x) = (root 2) / 2
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
dan815
  • dan815
|dw:1370823069333:dw|
dan815
  • dan815
the left side is a trig identity
Loser66
  • Loser66
good job, dan

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dan815
  • dan815
|dw:1370823187799:dw|
dan815
  • dan815
|dw:1370823217942:dw|
anonymous
  • anonymous
what about sec^2 (x) * tan((pi/2) - x) = 4
dan815
  • dan815
u can write tan(pi/2 - x) as sin ( pi/2-x) cos(x) ------------ = --------- cos(pi/2-x) sin(x)
anonymous
  • anonymous
so it would just be tan(x)?
dan815
  • dan815
no it will be 1/tan or cot
dan815
  • dan815
but forget that for now lets simplify sec
anonymous
  • anonymous
ohh, so then you're left with sec^2 x * cot x = 4
dan815
  • dan815
|dw:1370823441520:dw|
dan815
  • dan815
|dw:1370823530912:dw|
dan815
  • dan815
medal? :D
dan815
  • dan815
click best response
anonymous
  • anonymous
where'd that last step come from?
dan815
  • dan815
dividde by 2 and multipled denominator to right
dan815
  • dan815
|dw:1370823641656:dw|
dan815
  • dan815
like that
dan815
  • dan815
JUSSTT LIKE DAT THATS HOW I DOO IT
anonymous
  • anonymous
isn't it suppose to be sin(2x)?
dan815
  • dan815
|dw:1370823701785:dw|
dan815
  • dan815
oh wait are u sure.. lemme see i might have trig identity wrong
dan815
  • dan815
oh yes u are right.. is sin(2x)
dan815
  • dan815
|dw:1370823877158:dw|
dan815
  • dan815
there
anonymous
  • anonymous
the answers are suppose to be pi/12, 5pi/12, 13pi/12, and 17pi/12
dan815
  • dan815
pi/12 i just solved it ^
anonymous
  • anonymous
i understand the first 2 answers, where'd the other 2 come from?
dan815
  • dan815
click best response!
dan815
  • dan815
and i will teach u !
dan815
  • dan815
:D
anonymous
  • anonymous
what does clicking that do?
dan815
  • dan815
|dw:1370823983760:dw|
dan815
  • dan815
ok u get this step?
anonymous
  • anonymous
yeah
dan815
  • dan815
|dw:1370824010447:dw|
dan815
  • dan815
|dw:1370824027625:dw|
dan815
  • dan815
got it?
dan815
  • dan815
i just did those 2 steps at once before
anonymous
  • anonymous
there's suppose to be 4 answers in the domain of [0, 2pi)
dan815
  • dan815
okay then look where else this is true
dan815
  • dan815
|dw:1370824132653:dw|
dan815
  • dan815
only these 2 no more in domain 0 to 2pi unless u mean domain -2pi to 2pi
anonymous
  • anonymous
13pi/12 and 17pi/12 are also suppose to be solutions. I just don't know how to get them.
dan815
  • dan815
no those cant be right solution because that gives you negative 1/2
dan815
  • dan815
which question are u doing
anonymous
  • anonymous
sec^2 (x) tan ((pi/2) - x) = 4
anonymous
  • anonymous
the answer key from the back of my textbook says the answers are suppose to be pi/12, 5pi/12, 13pi/12, and 17pi/12
dan815
  • dan815
hmm i think u are right, 5pi/12 does work xD
dan815
  • dan815
http://www.wolframalpha.com/input/?i=1%2F2%3Dsin%282x%29#_=_
Loser66
  • Loser66
@dan815 do it by hand, please!!! he doesn't have math tool on test, dan
dan815
  • dan815
|dw:1370824635039:dw|
anonymous
  • anonymous
but here's the original problem http://www.wolframalpha.com/input/?i=sec^2+%28x%29+*+tan%28%28pi%2F2%29+-+x%29+%3D+4
anonymous
  • anonymous
somehow the method you showed me only got 2 of the 4 answers in the domain
dan815
  • dan815
@zepdrix solve this
dan815
  • dan815
sec^2 (x) tan ((pi/2) - x) = 4
Loser66
  • Loser66
1/2 = sin (2x) ---> 2x = sin^-1(1/2) = pi/6 ---> x = pi/12
anonymous
  • anonymous
pi/12 isn't the only answer though
Loser66
  • Loser66
sure, the reference angle only
anonymous
  • anonymous
arcsin (1/2) = pi/6 and 5pi/6
Loser66
  • Loser66
arcsin (1/2) =2x not =x , must divide both sides by 2 to get x
Loser66
  • Loser66
dan solve until sin(2x) =1/2
anonymous
  • anonymous
yeah, so the 2 answers are pi/12 and 5pi/12
dan815
  • dan815
|dw:1370824923788:dw|
dan815
  • dan815
there solve those other 2
dan815
  • dan815
iam just solving 1/2 = sin(x) u gotta divide by 2 to get your answers
anonymous
  • anonymous
where'd the 13pi/6 come from?
dan815
  • dan815
|dw:1370825262818:dw|
anonymous
  • anonymous
it's suppose to be one of the answers too
dan815
  • dan815
oh my goodness lets start over why am i getting confused lol
anonymous
  • anonymous
k
dan815
  • dan815
|dw:1370825389633:dw|
dan815
  • dan815
what are these 4 angles?
anonymous
  • anonymous
the vertical ones are for when cos x = 1/2, not when sin x = 1/2
anonymous
  • anonymous
sorry, i meant to say the opposite of that
dan815
  • dan815
oh ok what if the angle in the triangle
dan815
  • dan815
i guess that odesnt matter huh
anonymous
  • anonymous
oh wait, the first thing i said was right.
dan815
  • dan815
|dw:1370825681346:dw|
anonymous
  • anonymous
sin 2x = 1/2 -> 2x = arcsin (1/2) -> 2x = pi/6 or 5pi/6 -> x = pi/12 or 5pi/12
dan815
  • dan815
yeah
dan815
  • dan815
i dont know how theyre getting 2 more ...
dan815
  • dan815
what are the 2 more, i think i know xD
anonymous
  • anonymous
it appears that the function goes through 2 cycles between 0 and 2pi
anonymous
  • anonymous
http://www.wolframalpha.com/input/?i=sec^2+%28x%29+*+tan%28%28pi%2F2%29+-+x%29+%3D+4
anonymous
  • anonymous
look at the graph
dan815
  • dan815
|dw:1370825927771:dw|
dan815
  • dan815
now we know xD
dan815
  • dan815
|dw:1370825987989:dw|
anonymous
  • anonymous
YES!
dan815
  • dan815
ok so ur done?
anonymous
  • anonymous
those are the final answers, so how did you get the last 2?
dan815
  • dan815
i went around another roration
anonymous
  • anonymous
the 13pi/12 and 17pi/12?
dan815
  • dan815
|dw:1370826083546:dw|
anonymous
  • anonymous
how would you know to go around another rotation?
dan815
  • dan815
these are solution to sinx=1/2 but for sin(2x) = 1/2 these will become 13pi/12 and 17 pi / 12 so these are with in 0 to 2 pi now
anonymous
  • anonymous
oh yeah, period = 2pi/B
dan815
  • dan815
remember that n2pi + angle = another solution... for n = 0,1,2,3... infinity
dan815
  • dan815
|dw:1370826240191:dw|
anonymous
  • anonymous
how would you know to do that though?
dan815
  • dan815
what -.- okay
anonymous
  • anonymous
to add 2pi
dan815
  • dan815
solve sinx=1/2 forget the bounds what is the general solution?
anonymous
  • anonymous
general solution is pi/6 and 5pi/6
dan815
  • dan815
|dw:1370826296432:dw|
dan815
  • dan815
no this is the general solution
anonymous
  • anonymous
yup
dan815
  • dan815
okay so lets write the solutions for sin(2x) = 1/2 now that are with in 0 to 2 pi
dan815
  • dan815
|dw:1370826389012:dw|
dan815
  • dan815
so lets sub in n = 0, 1 ,2 ,3 and see where x still works
anonymous
  • anonymous
does that "2" in front of the x indicate that that's how many cycles there are in the period of 0 to 2pi?
dan815
  • dan815
|dw:1370826456275:dw|
dan815
  • dan815
|dw:1370826502259:dw|
dan815
  • dan815
for n=2, the number gets too big you can check it
dan815
  • dan815
so we stop here
anonymous
  • anonymous
okay got it, thanks
dan815
  • dan815
where do you live?
anonymous
  • anonymous
US
dan815
  • dan815
ok! study hard!! we need some more US ppl good at math =[
dan815
  • dan815
good luck!! I BELIEVE IN YOU!!

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