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