solve for x:
cos(x) - sin(x) = (root 2) / 2

- anonymous

solve for x:
cos(x) - sin(x) = (root 2) / 2

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- schrodinger

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- dan815

|dw:1370823069333:dw|

- dan815

the left side is a trig identity

- Loser66

good job, dan

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## More answers

- dan815

|dw:1370823187799:dw|

- dan815

|dw:1370823217942:dw|

- anonymous

what about sec^2 (x) * tan((pi/2) - x) = 4

- dan815

u can write tan(pi/2 - x) as
sin ( pi/2-x) cos(x)
------------ = ---------
cos(pi/2-x) sin(x)

- anonymous

so it would just be tan(x)?

- dan815

no it will be 1/tan or cot

- dan815

but forget that for now lets simplify sec

- anonymous

ohh, so then you're left with sec^2 x * cot x = 4

- dan815

|dw:1370823441520:dw|

- dan815

|dw:1370823530912:dw|

- dan815

medal? :D

- dan815

click best response

- anonymous

where'd that last step come from?

- dan815

dividde by 2 and multipled denominator to right

- dan815

|dw:1370823641656:dw|

- dan815

like that

- dan815

JUSSTT LIKE DAT THATS HOW I DOO IT

- anonymous

isn't it suppose to be sin(2x)?

- dan815

|dw:1370823701785:dw|

- dan815

oh wait are u sure.. lemme see i might have trig identity wrong

- dan815

oh yes u are right.. is sin(2x)

- dan815

|dw:1370823877158:dw|

- dan815

there

- anonymous

the answers are suppose to be pi/12, 5pi/12, 13pi/12, and 17pi/12

- dan815

pi/12 i just solved it ^

- anonymous

i understand the first 2 answers, where'd the other 2 come from?

- dan815

click best response!

- dan815

and i will teach u !

- dan815

:D

- anonymous

what does clicking that do?

- dan815

|dw:1370823983760:dw|

- dan815

ok u get this step?

- anonymous

yeah

- dan815

|dw:1370824010447:dw|

- dan815

|dw:1370824027625:dw|

- dan815

got it?

- dan815

i just did those 2 steps at once before

- anonymous

there's suppose to be 4 answers in the domain of [0, 2pi)

- dan815

okay then look where else this is true

- dan815

|dw:1370824132653:dw|

- dan815

only these 2 no more in domain 0 to 2pi unless u mean domain -2pi to 2pi

- anonymous

13pi/12 and 17pi/12 are also suppose to be solutions. I just don't know how to get them.

- dan815

no those cant be right solution because that gives you negative 1/2

- dan815

which question are u doing

- anonymous

sec^2 (x) tan ((pi/2) - x) = 4

- 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

hmm i think u are right, 5pi/12 does work xD

- dan815

http://www.wolframalpha.com/input/?i=1%2F2%3Dsin%282x%29#_=_

- Loser66

@dan815 do it by hand, please!!! he doesn't have math tool on test, dan

- dan815

|dw:1370824635039:dw|

- 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

somehow the method you showed me only got 2 of the 4 answers in the domain

- dan815

@zepdrix solve this

- dan815

sec^2 (x) tan ((pi/2) - x) = 4

- Loser66

1/2 = sin (2x) ---> 2x = sin^-1(1/2) = pi/6 ---> x = pi/12

- anonymous

pi/12 isn't the only answer though

- Loser66

sure, the reference angle only

- anonymous

arcsin (1/2) = pi/6 and 5pi/6

- Loser66

arcsin (1/2) =2x not =x , must divide both sides by 2 to get x

- Loser66

dan solve until sin(2x) =1/2

- anonymous

yeah, so the 2 answers are pi/12 and 5pi/12

- dan815

|dw:1370824923788:dw|

- dan815

there solve those other 2

- dan815

iam just solving 1/2 = sin(x)
u gotta divide by 2 to get your answers

- anonymous

where'd the 13pi/6 come from?

- dan815

|dw:1370825262818:dw|

- anonymous

it's suppose to be one of the answers too

- dan815

oh my goodness lets start over why am i getting confused lol

- anonymous

k

- dan815

|dw:1370825389633:dw|

- dan815

what are these 4 angles?

- anonymous

the vertical ones are for when cos x = 1/2, not when sin x = 1/2

- anonymous

sorry, i meant to say the opposite of that

- dan815

oh ok what if the angle in the triangle

- dan815

i guess that odesnt matter huh

- anonymous

oh wait, the first thing i said was right.

- dan815

|dw:1370825681346:dw|

- anonymous

sin 2x = 1/2 -> 2x = arcsin (1/2) -> 2x = pi/6 or 5pi/6 -> x = pi/12 or 5pi/12

- dan815

yeah

- dan815

i dont know how theyre getting 2 more ...

- dan815

what are the 2 more, i think i know xD

- anonymous

it appears that the function goes through 2 cycles between 0 and 2pi

- anonymous

http://www.wolframalpha.com/input/?i=sec^2+%28x%29+*+tan%28%28pi%2F2%29+-+x%29+%3D+4

- anonymous

look at the graph

- dan815

|dw:1370825927771:dw|

- dan815

now we know xD

- dan815

|dw:1370825987989:dw|

- anonymous

YES!

- dan815

ok so ur done?

- anonymous

those are the final answers, so how did you get the last 2?

- dan815

i went around another roration

- anonymous

the 13pi/12 and 17pi/12?

- dan815

|dw:1370826083546:dw|

- anonymous

how would you know to go around another rotation?

- 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

oh yeah, period = 2pi/B

- dan815

remember that n2pi + angle = another solution... for n = 0,1,2,3... infinity

- dan815

|dw:1370826240191:dw|

- anonymous

how would you know to do that though?

- dan815

what -.- okay

- anonymous

to add 2pi

- dan815

solve sinx=1/2 forget the bounds what is the general solution?

- anonymous

general solution is pi/6 and 5pi/6

- dan815

|dw:1370826296432:dw|

- dan815

no this is the general solution

- anonymous

yup

- dan815

okay so lets write the solutions for sin(2x) = 1/2 now that are with in 0 to 2 pi

- dan815

|dw:1370826389012:dw|

- dan815

so lets sub in n = 0, 1 ,2 ,3 and see where x still works

- 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

|dw:1370826456275:dw|

- dan815

|dw:1370826502259:dw|

- dan815

for n=2, the number gets too big you can check it

- dan815

so we stop here

- anonymous

okay got it, thanks

- dan815

where do you live?

- anonymous

US

- dan815

ok! study hard!! we need some more US ppl good at math =[

- dan815

good luck!! I BELIEVE IN YOU!!

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