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anonymous
 5 years ago
i need to solve these two problem using identities...sin2x+cos2x=0 and csc^2x(x/2)=2secx . help on either or both would be much appreciated!
anonymous
 5 years ago
i need to solve these two problem using identities...sin2x+cos2x=0 and csc^2x(x/2)=2secx . help on either or both would be much appreciated!

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0michelle i'm not sure if these identifies even exist.... i plugged in values for them and they don't work out. sin(2pi) + cos(2pi) = 1 though according to the ID you provided, it should be zero and the second..i'm not sure what it says is it csc(x/2)^(2x) = 2 secx?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0we're supposed to be solving for x using identities. sorry the second is confusing...its supposed to be\[ \csc ^{2}(x/2)=2secx\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0that should be an equal sign...it doesnt look like it on my computer, but it might just be screwy

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0okay michelle, i kinda gave up on the first and went straight to the second i really don't think that these exist because i've done it several ways, and I keep getting the same answer if i got different answers, i might be doing it wrong. but i get the same answer i'll do it out here and you can decide for yourself \[\csc^2 ({\frac{x}{2}}) = \frac{2}{\cos x}\] \[\frac{1}{\sin ^2 (\frac{x}{2})} = \frac{2}{\cos x}\] \[2\sin^2(\frac{x}{2}) = \cos x\] \[2(\frac{1  \cos x}{2}) = \cos x\] \[1  \cos x = \cos x\] \[1 = 2 \cos x\] \[\cos x = \frac{1}{2}\] ^ i keep getting that TT

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0thats what we're supposed to get! thanks so much...that helps a lot. we're solving for x. and i can solve for cosx=1/2, because we know that x=pi/3 or 5pi/3. thats what we're supposed to be doing =)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0AHHHH MICHELLE!!! THAT MEANS THESE AREN'T IDENTITIES!!! RAWRRRRRRRRRRRRRRR you had me tearing my hair out trying to get the equation to equal.. TT sighs give me a bit more time, i'll do the first one

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i'll try doing the first one*** i'm not sure if i'll be able to

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0omg im so sorry. i meant we're solving for x like, using the different identities. idk...thats just what it says on the top of my sheet..."solve for x usiing identities". im sorry...you really dont have to. thanks for that one tho =)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0lol so whos up to what here ?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i got the second one see if you can do the first one solve for x

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yeh , its relatively easy lol

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0what i've got so far, i think is right \[2 \sin x * \cos x + \cos^2 x  \sin^2 x = 0\] \[2 \tan + 1  \tan^2 x = 0\] \[\tan^2 x  2 \tan x  1 = 0\] now you have to solve for x using the quadratic formula, which i dont know how to take into account the tangent

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you made a mistake between the 2nd last and the last line a +1 becoame a 1

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0no, i negated the entire thing, it is NOT a perfect square

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you are doing it the hard way anyway

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0wait im confused. how did you get from the first step to the second step (with the tans)? are there identities that tell you that? i might just not know them

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0then, please, tell us how to do it...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i divided the entire equation by sin^2(x)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0sin(2x) = cos(2x) tan(2x) =1 related angle of pi/4 tan is negative in the 2nd and 4th quadrants now we originally had 0<=x<=2pi , multiply through by 2 ( because we have (2x) as the argument of our tan )

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so 0<=x<=4pi ( ie we need two revolutions

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0trig id overboard.... michelle, listen to elec, he's right

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so 2x = 3pi/4 , 7pi/4 ( from the first revolution )

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0now, we need another revolution, so add 2pi to each of these

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so 2x = 3pi/4 , 7pi/4 , 11pi/ 4 , 15pi/ 4 then divide by 2

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0x = \[x = (3\pi/8 ) , (7\pi/8) , (11\pi/8), (15\pi/8) \]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0wow. thank you so so so much. you guys are both like, geniuses. thanks =)) i gave you both medals, whatever that means.
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